TnT/Multi_CodeNet4Repair huggingface dataset:
Problem: We have a tree with N vertices. The i-th edge connects Vertex A_i and B_i bidirectionally.  Takahashi is standing at Vertex u, and Aoki is standing at Vertex v.  Now, they will play a game of tag as follows:    * 1. If Takahashi and Aoki are standing at the same vertex, the game ends. Otherwise, Takahashi moves to a vertex of his choice that is adjacent to his current vertex.    * 2. If Takahashi and Aoki are standing at the same vertex, the game ends. Otherwise, Aoki moves to a vertex of his choice that is adjacent to his current vertex.    * 3. Go back to step 1.  Takahashi performs his moves so that the game ends as late as possible, while Aoki performs his moves so that the game ends as early as possible.  Find the number of moves Aoki will perform before the end of the game if both Takahashi and Aoki know each other's position and strategy.  It can be proved that the game is bound to end.

Submission attempts (in order):
N,u,v = map(int,input().split())
edge = [[] for i in range(N)]
for i in range(N-1):
  a,b = map(int, input().split())
  edge[a-1].append(b-1)
  edge[b-1].append(a-1)

def bfs(s):
  dist = [-1]*N
  que = [s-1]
  while que:
    v = que.pop()
    d = dist[v]
    for w in edge[v]:
      if dist[w]>-1: continue
      dist[w] = d+1
      que.append(w)
  return dist

ud,vd,ans = bfs(u),bfs(v),0
for ud_i,vd_i in zip(ud,vd): ans = max(ans, vd_i-1) if ud_i<vd_i else ans
print(ans)

from collections import deque
N,u,v = map(int,input().split())
edge = [[] for i in range(N)]
for i in range(N-1):
  a,b = map(int, input().split())
  edge[a-1].append(b-1)
  edge[b-1].append(a-1)

def bfs(s):
  dist = [-1]*N
  que = deque([s-1])
  dist[s-1] = 0
  while que:
    v = que.popleft()
    d = dist[v]
    for w in edge[v]:
      if dist[w]>-1: continue
      dist[w] = d+1
      que.append(w)
  return dist

ud,vd = bfs(u),bfs(v)
ans = 0
for ud_i,vd_i in zip(ud,vd): ans = max(ans, vd_i-1) if ud_i<vd_i else ans
print(ans)


Verdicts: Wrong Answer

Accepted
Problem: We have a tree with N vertices. The i-th edge connects Vertex A_i and B_i bidirectionally.  Takahashi is standing at Vertex u, and Aoki is standing at Vertex v.  Now, they will play a game of tag as follows:    * 1. If Takahashi and Aoki are standing at the same vertex, the game ends. Otherwise, Takahashi moves to a vertex of his choice that is adjacent to his current vertex.    * 2. If Takahashi and Aoki are standing at the same vertex, the game ends. Otherwise, Aoki moves to a vertex of his choice that is adjacent to his current vertex.    * 3. Go back to step 1.  Takahashi performs his moves so that the game ends as late as possible, while Aoki performs his moves so that the game ends as early as possible.  Find the number of moves Aoki will perform before the end of the game if both Takahashi and Aoki know each other's position and strategy.  It can be proved that the game is bound to end.

Submission attempts (in order):
from collections import deque, defaultdict
from copy import deepcopy

N ,u, v = map(int,input().split())
graph = defaultdict(deque)
for _ in range(1,N):
    a, b = map(int,input().split())
    graph[a].append(b)
    graph[b].append(a)

depth = dict(zip(list(range(1,N+1)),[-1]*N))

def bfs(s):
    if graph[s]:
        for v in graph[s]:
            if depth[v] == -1:
                depth[v] = depth[s] + 1
                bfs(v)

depth = dict(zip(list(range(1,N+1)),[-1]*N))
depth[v] = 0
bfs(v)
depth_v = deepcopy(depth)

depth = dict(zip(list(range(1,N+1)),[-1]*N))
depth[u] = 0
bfs(u)
depth_u = depth
depth_v = sorted(depth_v.items(), key = lambda x:x[0])
depth_u = sorted(depth_u.items(), key= lambda x:x[0])
print(depth_v)
print(depth_u)
MAX = 0
for v,u in zip(depth_v, depth_u):
    print(v[1],u[1])
    if u[1] < v[1]:
        MAX = max(MAX,v[1])
print(MAX-1)

from collections import deque,defaultdict
import copy

def main():
    from collections import defaultdict
    import copy 
    n,u,v = map(int,input().split())
    u -= 1
    v -= 1
    graph = defaultdict(deque)
    for _ in range(n-1):
        a,b = map(int,input().split())
        a -= 1
        b -= 1
        graph[a].append(b)
        graph[b].append(a)

    depth = dfs(graph,u,n)
    # print(depth)
    depth2 = dfs(graph,v,n)
    # print(depth2)
    ans = 0
    ans2 = 10**6
    for i,j in zip(depth,depth2):
        if i < j:
            ans = max(j,ans)
        
        #     ans2 = min(j,ans2)

    print(ans-1)

def dfs(G,s,n):
    stack = deque([s])
    graph = copy.deepcopy(G)
    depth = [None] * n
    depth[s] = 0
    while stack:
        u = stack[-1]
        if graph[u]:
            v = graph[u].popleft()
            if depth[v] == None:
                depth[v] = depth[u] + 1
                stack.append(v)

        else:
            stack.pop()

    return depth
    

main()

Verdicts: Runtime Error

Accepted
Problem: We have a tree with N vertices. The i-th edge connects Vertex A_i and B_i bidirectionally.  Takahashi is standing at Vertex u, and Aoki is standing at Vertex v.  Now, they will play a game of tag as follows:    * 1. If Takahashi and Aoki are standing at the same vertex, the game ends. Otherwise, Takahashi moves to a vertex of his choice that is adjacent to his current vertex.    * 2. If Takahashi and Aoki are standing at the same vertex, the game ends. Otherwise, Aoki moves to a vertex of his choice that is adjacent to his current vertex.    * 3. Go back to step 1.  Takahashi performs his moves so that the game ends as late as possible, while Aoki performs his moves so that the game ends as early as possible.  Find the number of moves Aoki will perform before the end of the game if both Takahashi and Aoki know each other's position and strategy.  It can be proved that the game is bound to end.

Submission attempts (in order):
import sys
sys.setrecursionlimit(200000)

def dfs(array, current, count):
  array[current - 1] = count
  for neighbour in graph[current]:
    if array[neighbour - 1] == -1:
      array = dfs(array, neighbour, count + 1)
  else:
    return array

T = [-1 for x in range(100000)]
A = [-1 for x in range(100000)]

graph = {}

N,u,v = map(int,input().split())

for i in range(N-1):
  a,b = map(int,input().split())
  try:
    graph[a].append(b)
  except KeyError:
    graph[a] = [b]

  try:
    graph[b].append(a)
  except KeyError:
    graph[b] = [a]

T = dfs(T, u, 0)
A = dfs(A, v, 0)

ans = -1
for i in range(N):
  if T[i] < A[i]:
    ans = max(ans, A[ImportError] - 1)
print(max(0,ans))


import sys
sys.setrecursionlimit(200000)

def dfs(array, current, count):
  array[current - 1] = count
  for neighbour in graph[current]:
    if array[neighbour - 1] == -1:
      array = dfs(array, neighbour, count + 1)
  else:
    return array

T = [-1 for x in range(100000)]
A = [-1 for x in range(100000)]

graph = {}

N,u,v = map(int,input().split())

for i in range(N-1):
  a,b = map(int,input().split())
  try:
    graph[a].append(b)
  except KeyError:
    graph[a] = [b]

  try:
    graph[b].append(a)
  except KeyError:
    graph[b] = [a]

T = dfs(T, u, 0)
A = dfs(A, v, 0)

ans = -1
for i in range(N):
  if T[i] < A[i]:
    ans = max(ans, A[i] - 1)
print(max(0,ans))


Verdicts: Runtime Error

Accepted
Problem: We have a tree with N vertices. The i-th edge connects Vertex A_i and B_i bidirectionally.  Takahashi is standing at Vertex u, and Aoki is standing at Vertex v.  Now, they will play a game of tag as follows:    * 1. If Takahashi and Aoki are standing at the same vertex, the game ends. Otherwise, Takahashi moves to a vertex of his choice that is adjacent to his current vertex.    * 2. If Takahashi and Aoki are standing at the same vertex, the game ends. Otherwise, Aoki moves to a vertex of his choice that is adjacent to his current vertex.    * 3. Go back to step 1.  Takahashi performs his moves so that the game ends as late as possible, while Aoki performs his moves so that the game ends as early as possible.  Find the number of moves Aoki will perform before the end of the game if both Takahashi and Aoki know each other's position and strategy.  It can be proved that the game is bound to end.

Submission attempts (in order):
from collections import deque

n, s, t = map(int, input().split(' '))
s -= 1
t -= 1

v = [[] for i in range(n)]
f = [0]*n

for i in range(n-1):
    a, b = map(int, input().split(' '))
    a -= 1
    b -= 1
    v[a].append(b)
    v[b].append(a)
    f[a] += 1
    f[b] += 1
for i in range(n):
    if f[i] == 0:
        l.append(i)
depth = [-1]*n
depth[t] = 0
q = deque([t])
while q:
    x = q.popleft()
    for y in v[x]:
        if depth[y] < 0:
            depth[y] = depth[x]+1
            q.append(y)
l = depth[:]
d = [-1]*n
d[s] = 0
q = deque([s])
while q:
    x = popleft()
    for y in v[x]:
        if d[y] < 0:
            d[y] = d[x]+1
            q.append(y)
l = sorted(range(n), key=lambda x:-depth[x])
for i in l:
    if d[i] < depth[i]:
        break
print(depth[i] - 1)



from collections import deque

n, s, t = map(int, input().split(' '))
s -= 1
t -= 1

v = [[] for i in range(n)]
f = [0]*n

for i in range(n-1):
    a, b = map(int, input().split(' '))
    a -= 1
    b -= 1
    v[a].append(b)
    v[b].append(a)
    f[a] += 1
    f[b] += 1
for i in range(n):
    if f[i] == 0:
        l.append(i)
depth = [-1]*n
depth[t] = 0
q = deque([t])
while q:
    x = q.popleft()
    for y in v[x]:
        if depth[y] < 0:
            depth[y] = depth[x]+1
            q.append(y)
l = depth[:]
d = [-1]*n
d[s] = 0
q = deque([s])
while q:
    x = q.popleft()
    for y in v[x]:
        if d[y] < 0:
            d[y] = d[x]+1
            q.append(y)
l = sorted(range(n), key=lambda x:-depth[x])
for i in l:
    if d[i] < depth[i]:
        break
print(depth[i] - 1)



Verdicts: Runtime Error

Accepted
Problem: We have a tree with N vertices. The i-th edge connects Vertex A_i and B_i bidirectionally.  Takahashi is standing at Vertex u, and Aoki is standing at Vertex v.  Now, they will play a game of tag as follows:    * 1. If Takahashi and Aoki are standing at the same vertex, the game ends. Otherwise, Takahashi moves to a vertex of his choice that is adjacent to his current vertex.    * 2. If Takahashi and Aoki are standing at the same vertex, the game ends. Otherwise, Aoki moves to a vertex of his choice that is adjacent to his current vertex.    * 3. Go back to step 1.  Takahashi performs his moves so that the game ends as late as possible, while Aoki performs his moves so that the game ends as early as possible.  Find the number of moves Aoki will perform before the end of the game if both Takahashi and Aoki know each other's position and strategy.  It can be proved that the game is bound to end.

Submission attempts (in order):
def dps(no, fno, d, hen, D, MD, P):
    D[no] = d
    P[no] = fno
    maxd = d
    for h in hen[no]:
        if h == fno:
            continue
        maxd = max(maxd, dps(h, no, d + 1, hen, D, MD, P))
    MD[no] = maxd
    return maxd


def main():
    import my_method
    N, u, v = map(int, my_method.input().split())
    hen = {}
    for i in range(N - 1):
        A, B = map(int, my_method.input().split())
        if A in hen.keys():
            hen[A].append(B)
        else:
            hen[A] = [B]
        if B in hen.keys():
            hen[B].append(A)
        else:
            hen[B] = [A]
    DEPTH = [-1 for i in range(N + 1)]
    MAX_DEPTH = [-1 for i in range(N + 1)]
    Parent = [-1 for i in range(N + 1)]
    dps(v, -1, 0, hen, DEPTH, MAX_DEPTH, Parent)
    ans = DEPTH[u]
    p = u
    for i in range((DEPTH[u] + 1) // 2 - 1):
        p = Parent[p]
        ans = max(ans, MAX_DEPTH[p])
    print(ans - 1)

if __name__ == '__main__':
    main()


def dps_rec(no, fno, d, hen, D, MD, P):
    D[no] = d
    P[no] = fno
    maxd = d
    for h in hen[no]:
        if h == fno:
            continue
        maxd = max(maxd, dps_rec(h, no, d + 1, hen, D, MD, P))
    MD[no] = maxd
    return maxd

def dps_stc(no, fno, d, hen, D):
    stack = [(no, fno, d)]
    while not stack == []:
        t = stack.pop()
        D[t[0]] = t[2]
        for h in hen[t[0]]:
            if h == t[1]:
                continue
            stack.append((h, t[0], t[2] + 1))

def main():
    N, u, v = map(int, input().split())
    hen = {}
    for i in range(N - 1):
        A, B = map(int, input().split())
        if A in hen.keys():
            hen[A].append(B)
        else:
            hen[A] = [B]
        if B in hen.keys():
            hen[B].append(A)
        else:
            hen[B] = [A]
    DEPTH_A = [-1 for i in range(N + 1)]
    DEPTH_T = [-1 for i in range(N + 1)]
    dps_stc(v, -1, 0, hen, DEPTH_A)
    dps_stc(u, -1, 0, hen, DEPTH_T)
    ans = 0
    for a, t in zip(DEPTH_A, DEPTH_T):
        if a > t:
            ans = max(ans, a - 1)

    print(ans)


if __name__ == '__main__':
    main()


Verdicts: Runtime Error

Accepted
Problem: We have a tree with N vertices. The i-th edge connects Vertex A_i and B_i bidirectionally.  Takahashi is standing at Vertex u, and Aoki is standing at Vertex v.  Now, they will play a game of tag as follows:    * 1. If Takahashi and Aoki are standing at the same vertex, the game ends. Otherwise, Takahashi moves to a vertex of his choice that is adjacent to his current vertex.    * 2. If Takahashi and Aoki are standing at the same vertex, the game ends. Otherwise, Aoki moves to a vertex of his choice that is adjacent to his current vertex.    * 3. Go back to step 1.  Takahashi performs his moves so that the game ends as late as possible, while Aoki performs his moves so that the game ends as early as possible.  Find the number of moves Aoki will perform before the end of the game if both Takahashi and Aoki know each other's position and strategy.  It can be proved that the game is bound to end.

Submission attempts (in order):
print("123".replace(input(),"").replace(input(),""))

n, u, v, *L = map(int, open(0).read().split())
dist_f_v = [-1] * n
con = [[] for _ in range(n)]
for s, t in zip(*[iter(L)] * 2):
	con[s - 1].append(t - 1)
	con[t - 1].append(s - 1)
dist_f_v[v - 1] = 0
q = [v - 1]
while q:
	cur = q.pop()
	for nxt in con[cur]:
		if dist_f_v[nxt] < 0:
			dist_f_v[nxt] = dist_f_v[cur] + 1
			q.append(nxt)
dist_f_u = [-1] * n
dist_f_u[u - 1] = 0
q = [u - 1]
m = 0
while q:
	cur = q.pop()
	m = max(m, dist_f_v[cur])
	for nxt in con[cur]:
		if dist_f_u[nxt] > -1:
			continue
		if dist_f_u[cur] + 1 < dist_f_v[nxt]:
			dist_f_u[nxt] = dist_f_u[cur] + 1
			q.append(nxt)
		elif dist_f_u[cur] + 1 == dist_f_v[nxt]:
			dist_f_u[nxt] = dist_f_u[cur] + 1
print(m - 1)


Verdicts: Wrong Answer

Accepted
Problem: We have a tree with N vertices. The i-th edge connects Vertex A_i and B_i bidirectionally.  Takahashi is standing at Vertex u, and Aoki is standing at Vertex v.  Now, they will play a game of tag as follows:    * 1. If Takahashi and Aoki are standing at the same vertex, the game ends. Otherwise, Takahashi moves to a vertex of his choice that is adjacent to his current vertex.    * 2. If Takahashi and Aoki are standing at the same vertex, the game ends. Otherwise, Aoki moves to a vertex of his choice that is adjacent to his current vertex.    * 3. Go back to step 1.  Takahashi performs his moves so that the game ends as late as possible, while Aoki performs his moves so that the game ends as early as possible.  Find the number of moves Aoki will perform before the end of the game if both Takahashi and Aoki know each other's position and strategy.  It can be proved that the game is bound to end.

Submission attempts (in order):
n, uplus, vplus = map(int, input().split())
u, v = uplus-1, vplus-1
adj = [[] for _ in range(n)]
for _ in range(n-1):
  aplus, bplus = map(int, input().split())
  a, b = aplus-1, bplus-1
  adj[a].append(b)
  adj[b].append(a)

u_dist = [n] * n
u_dist[u] = 0
stack = [u]
while stack:
  vertex = stack.pop()
  for nxt in adj[vertex]:
    if u_dist[nxt] == n:
      u_dist[nxt] = u_dist[vertex] + 1
      stack.append(nxt)

v_dist = [n] * n
v_dist[v] = 0
stack = [v]
leaves = []
while stack:
  vertex = stack.pop()
  isleaf = True
  for nxt in adj[vertex]:
    if u_dist[nxt] == n:
      u_dist[nxt] = u_dist[vertex] + 1
      stack.append(nxt)
      isleaf = False
  if isleaf:
    leaves.append(vertex)

max_depth = 0
for leaf in leaves:
  depth = v_dist[leaf]
  if u_dist[leaf] < depth and max_depth < depth:
    max_depth = depth
    
print(max_depth - 1)

n, uplus, vplus = map(int, input().split())
u, v = uplus-1, vplus-1
adj = [[] for _ in range(n)]
for _ in range(n-1):
  aplus, bplus = map(int, input().split())
  a, b = aplus-1, bplus-1
  adj[a].append(b)
  adj[b].append(a)

u_dist = [n] * n
u_dist[u] = 0
stack = [u]
while stack:
  vertex = stack.pop()
  for nxt in adj[vertex]:
    if u_dist[nxt] == n:
      u_dist[nxt] = u_dist[vertex] + 1
      stack.append(nxt)

v_dist = [n] * n
v_dist[v] = 0
stack = [v]
leaves = []
while stack:
  vertex = stack.pop()
  isleaf = True
  for nxt in adj[vertex]:
    if v_dist[nxt] == n:
      v_dist[nxt] = v_dist[vertex] + 1
      stack.append(nxt)
      isleaf = False
  if isleaf:
    leaves.append(vertex)

max_depth = 0
for leaf in leaves:
  depth = v_dist[leaf]
  if u_dist[leaf] < depth and max_depth < depth:
    max_depth = depth
    
print(max_depth - 1)

Verdicts: Wrong Answer

Accepted
Problem: We have a tree with N vertices. The i-th edge connects Vertex A_i and B_i bidirectionally.  Takahashi is standing at Vertex u, and Aoki is standing at Vertex v.  Now, they will play a game of tag as follows:    * 1. If Takahashi and Aoki are standing at the same vertex, the game ends. Otherwise, Takahashi moves to a vertex of his choice that is adjacent to his current vertex.    * 2. If Takahashi and Aoki are standing at the same vertex, the game ends. Otherwise, Aoki moves to a vertex of his choice that is adjacent to his current vertex.    * 3. Go back to step 1.  Takahashi performs his moves so that the game ends as late as possible, while Aoki performs his moves so that the game ends as early as possible.  Find the number of moves Aoki will perform before the end of the game if both Takahashi and Aoki know each other's position and strategy.  It can be proved that the game is bound to end.

Submission attempts (in order):
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
sys.setrecursionlimit(10 ** 7)

from collections import deque

n, u, v = map(int, readline().split())
graph = [[] for _ in range(n + 1)]
for i in range(n - 1):
    a, b = map(int, readline().split())
    graph[a].append(b)
    graph[b].append(a)


def dfs(v):
    dist = [-1] * (n + 1)
    stack = deque([v])
    dist[v] = 0
    while stack:
        v = stack.popleft()
        memo = dist[v] + 1
        for w in graph[v]:
            if dist[w] >= 0:
                continue
            dist[w] = memo
            stack.append(w)
    return dist


U, V = dfs(u), dfs(v)
ans = 0
for x, y in zip(U[1:], V[1:]):
    print(x, y)
    if x < y:
        ans = max(ans, y - 1)
print(ans)


import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
sys.setrecursionlimit(10 ** 7)

from collections import deque

n, u, v = map(int, readline().split())
graph = [[] for _ in range(n + 1)]
for i in range(n - 1):
    a, b = map(int, readline().split())
    graph[a].append(b)
    graph[b].append(a)


def dfs(v):
    dist = [-1] * (n + 1)
    stack = deque([v])
    dist[v] = 0
    while stack:
        v = stack.popleft()
        memo = dist[v] + 1
        for w in graph[v]:
            if dist[w] >= 0:
                continue
            dist[w] = memo
            stack.append(w)
    return dist


U, V = dfs(u), dfs(v)
ans = 0
for x, y in zip(U[1:], V[1:]):
    if x < y:
        ans = max(ans, y - 1)
print(ans)


Verdicts: Wrong Answer

Accepted
Problem: We have a tree with N vertices. The i-th edge connects Vertex A_i and B_i bidirectionally.  Takahashi is standing at Vertex u, and Aoki is standing at Vertex v.  Now, they will play a game of tag as follows:    * 1. If Takahashi and Aoki are standing at the same vertex, the game ends. Otherwise, Takahashi moves to a vertex of his choice that is adjacent to his current vertex.    * 2. If Takahashi and Aoki are standing at the same vertex, the game ends. Otherwise, Aoki moves to a vertex of his choice that is adjacent to his current vertex.    * 3. Go back to step 1.  Takahashi performs his moves so that the game ends as late as possible, while Aoki performs his moves so that the game ends as early as possible.  Find the number of moves Aoki will perform before the end of the game if both Takahashi and Aoki know each other's position and strategy.  It can be proved that the game is bound to end.

Submission attempts (in order):
import sys
sys.recursionlimit(10**5)

N,u,v=map(int,input().split())

du,dv=[0]*-~N,[0]*-~N
T=[list() for i in range(-~N)]

def dfs(p,t,v,d):
  d[v]=t
  for i in T[v]:
    if i!=p:
      dfs(v,t+1,i,d)

for i in range(N-1):
  a,b=map(int,input().split())
  T[a].append(b)
  T[b].append(a)

dfs(0,0,u,du)
dfs(0,0,v,dv)

c=0
for i in range(1,-~N):
  if len(T[i])==1 and i!=v:
    if du[i]<dv[i]:
      c=max(c,~-dv[i])

print(c)


N,u,v=map(int,input().split())

du,dv=[0]*-~N,[0]*-~N
T=[list() for i in range(-~N)]

def dsf(p,t,v,d):
  d[v]=t
  for i in T[v]:
    if i!=p:
      dsf(v,t+1,i,d)

for i in range(N-1):
  a,b=map(int,input().split())
  T[a].append(b)
  T[b].append(a)

dsf(0,0,u,du)
dsf(0,0,v,dv)

if du[v]==1 and len(T[v])==1:
  print(0)
else:
  m=0
  for i in range(1,~-N):
    if du[i] < dv[i]:
      if m<du[i]:
        v,m=i,du[i]

  if m!=0:
    print(du[v]+m-1)
  else:
    m=N+1
    for i in range(1,~-N):
      if du[i]==dv[i]:
        m=min(m,du[i])
    print(m)


import sys
sys.setrecursionlimit(10**5)
N,u,v=map(int,input().split())

du,dv=[0]*-~N,[0]*-~N
T=[list() for i in range(-~N)]

def dfs(p,t,v,d):
  d[v]=t
  for i in T[v]:
    if i!=p:
      dfs(v,t+1,i,d)

for i in range(N-1):
  a,b=map(int,input().split())
  T[a].append(b)
  T[b].append(a)

dfs(0,0,u,du)
dfs(0,0,v,dv)

c=0
for i in range(1,-~N):
  if len(T[i])==1 and i!=v:
    if du[i]<dv[i]:
      c=max(c,~-dv[i])

print(c)



Verdicts: Runtime Error

Runtime Error

Accepted
Problem: We have a tree with N vertices. The i-th edge connects Vertex A_i and B_i bidirectionally.  Takahashi is standing at Vertex u, and Aoki is standing at Vertex v.  Now, they will play a game of tag as follows:    * 1. If Takahashi and Aoki are standing at the same vertex, the game ends. Otherwise, Takahashi moves to a vertex of his choice that is adjacent to his current vertex.    * 2. If Takahashi and Aoki are standing at the same vertex, the game ends. Otherwise, Aoki moves to a vertex of his choice that is adjacent to his current vertex.    * 3. Go back to step 1.  Takahashi performs his moves so that the game ends as late as possible, while Aoki performs his moves so that the game ends as early as possible.  Find the number of moves Aoki will perform before the end of the game if both Takahashi and Aoki know each other's position and strategy.  It can be proved that the game is bound to end.

Submission attempts (in order):
import sys
input = sys.stdin.readline
 

sys.setrecursionlimit(10**6)

n, u, v = map(int, input().split())
u, v = u-1, v-1
l = [None]*n
for i in range(n-1):
  a, b = map(int, input().split())
  a, b = a-1, b-1
  l[a].append(b)
  l[b].append(a)

def dfs(G, cr, seen, dist):
  seen[cr] = 1
  for i in range(G[cr]):
    if seen[i] == 0:
      dist[i] = dist[cr] + 1
      dfs(G, i, seen, dist)
  return dist

seen_u = [0]*n
seen_v = [0]*n
dist_u = [0]*n
dist_v = [0]*n

T1 = dfs(l, u, seen_u, dist_u)
T2 = dfs(l, v, seen_v, dist_v)

for i in range(n):
  if dist_u[i] >= dist_v[i]:
    dist_v[i] = -1
    
print(max(dist_v)-1)

import sys
input = sys.stdin.readline
 

sys.setrecursionlimit(10**6)

n, u, v = map(int, input().split())
u, v = u-1, v-1
l = [[] for i in range(n)]
for i in range(n-1):
  a, b = map(int, input().split())
  a, b = a-1, b-1
  l[a].append(b)
  l[b].append(a)

def dfs(G, cr, seen, dist):
  seen[cr] = 1
  for i in G[cr]:
    if seen[i] == 0:
      dist[i] = dist[cr] + 1
      dfs(G, i, seen, dist)
  return dist

seen_u = [0]*n
seen_v = [0]*n
dist_u = [0]*n
dist_v = [0]*n

T1 = dfs(l, u, seen_u, dist_u)
T2 = dfs(l, v, seen_v, dist_v)

for i in range(n):
  if dist_u[i] >= dist_v[i]:
    dist_v[i] = -1
    
print(max(dist_v)-1)

Verdicts: Runtime Error

Accepted
Problem: We have a tree with N vertices. The i-th edge connects Vertex A_i and B_i bidirectionally.  Takahashi is standing at Vertex u, and Aoki is standing at Vertex v.  Now, they will play a game of tag as follows:    * 1. If Takahashi and Aoki are standing at the same vertex, the game ends. Otherwise, Takahashi moves to a vertex of his choice that is adjacent to his current vertex.    * 2. If Takahashi and Aoki are standing at the same vertex, the game ends. Otherwise, Aoki moves to a vertex of his choice that is adjacent to his current vertex.    * 3. Go back to step 1.  Takahashi performs his moves so that the game ends as late as possible, while Aoki performs his moves so that the game ends as early as possible.  Find the number of moves Aoki will perform before the end of the game if both Takahashi and Aoki know each other's position and strategy.  It can be proved that the game is bound to end.

Submission attempts (in order):
from sys import setrecursionlimit
setrecursionlimit(10101)

N, u, v = map(int, input().split())
u -= 1
v -= 1
table = [[] for _ in range(N)]
for _ in range(N-1):
    a, b = map(int, input().split())
    a -= 1
    b -= 1
    table[a].append(b)
    table[b].append(a)

depthu = [-1] * (N)
depthu[u] = 0
depthv = [-1] * (N)
depthv[v] = 0

def dfs_depth(p, depth):
    for q in table[p]:
        if depth[q] == -1:
            depth[q] = depth[p] + 1
            dfs_depth(q, depth)

dfs_depth(v, depthv)
dfs_depth(u, depthu)

p = [i for i in range(1, N-1) if depthu[i] >= depthv[i] + 2]
if p:
    print(max(depthu[i] + (depthu[i] - depthv[i]) // 2 for i in p))
else:
    print(depthv[u] // 2)


from collections import deque

N, u, v = map(int, input().split())
u -= 1
v -= 1
table = [[] for _ in range(N)]
for _ in range(N-1):
    a, b = map(int, input().split())
    a -= 1
    b -= 1
    table[a].append(b)
    table[b].append(a)

depthu = [-1] * N
depthu[u] = 0
depthv = [-1] * N
depthv[v] = 0

par = deque([u])
while par:
    p = par.popleft()
    for q in table[p]:
        if depthu[q] == -1:
            depthu[q] = depthu[p] + 1
            par.append(q)

par = deque([v])
while par:
    p = par.popleft()
    for q in table[p]:
        if depthv[q] == -1:
            depthv[q] = depthv[p] + 1
            par.append(q)

p = [i for i in range(N) if depthu[i] < depthv[i]]
if p:
    print(max(depthv[i] - 1 for i in p))
else:
    print(depthu[v] - 1)

Verdicts: Runtime Error

Accepted
Problem: We have a tree with N vertices. The i-th edge connects Vertex A_i and B_i bidirectionally.  Takahashi is standing at Vertex u, and Aoki is standing at Vertex v.  Now, they will play a game of tag as follows:    * 1. If Takahashi and Aoki are standing at the same vertex, the game ends. Otherwise, Takahashi moves to a vertex of his choice that is adjacent to his current vertex.    * 2. If Takahashi and Aoki are standing at the same vertex, the game ends. Otherwise, Aoki moves to a vertex of his choice that is adjacent to his current vertex.    * 3. Go back to step 1.  Takahashi performs his moves so that the game ends as late as possible, while Aoki performs his moves so that the game ends as early as possible.  Find the number of moves Aoki will perform before the end of the game if both Takahashi and Aoki know each other's position and strategy.  It can be proved that the game is bound to end.

Submission attempts (in order):
2


N, u, v = map(int, input().split())
u -= 1
v -= 1

e_list = [[] for _ in range(N)]

for _ in range(N - 1):
	a, b = map(int, input().split())
	a -= 1
	b -= 1
	e_list[a].append(b)
	e_list[b].append(a)

def calc_depth_table(v):
	queue = [[v, None, 0]]
	depth_table = [0] * N
	depth_table[v] = 0
	
	while queue:
		cur, pre, depth = queue.pop()
		depth += 1
		for n in e_list[cur]:
			if n != pre:
				queue.append([n, cur, depth])
				depth_table[n] = depth
	return depth_table

def calc_depth_table2(depth_from_v, u):
	queue = [[u, None, 0, depth_from_v[u], u]]
	depth_table = [0] * N
	depth_table[u] = (0, depth_from_v[u], u)
	
	while queue:
		cur, pre, depth, min_depth_from_v, min_depth_point = queue.pop()
		depth += 1
		for n in e_list[cur]:
			if n != pre:
				depth2 = depth_from_v[n]
				if min_depth_from_v > depth2:
					new_min_depth = depth2
					new_min_depth_point = n
				else:
					new_min_depth = min_depth_from_v
					new_min_depth_point = min_depth_point
				queue.append([n, cur, depth, new_min_depth, new_min_depth_point])
				depth_table[n] = (depth, new_min_depth, new_min_depth_point)
	return depth_table

depth_from_v = calc_depth_table(v)
depth_from_u = calc_depth_table2(depth_from_v, u)

#print(depth_from_v)
#print(depth_from_u)

max_count = 0
for i in range(N):
	a_dist = depth_from_v[i]
	t_dist, a_hit_dist, hit_pos = depth_from_u[i]
	t_hit_dist = depth_from_u[hit_pos][0]
	
#	print(t_hit_dist, a_hit_dist)
	if t_hit_dist < a_hit_dist:
#		print(t_hit_dist, a_hit_dist)
#		if (a_hit_dist - t_hit_dist) % 2 == 1:
		if max_count < a_dist - 1:
			max_count = a_dist - 1
#			else:
#				

print(max_count)


Verdicts: Wrong Answer

Accepted
Problem: We have a tree with N vertices. The i-th edge connects Vertex A_i and B_i bidirectionally.  Takahashi is standing at Vertex u, and Aoki is standing at Vertex v.  Now, they will play a game of tag as follows:    * 1. If Takahashi and Aoki are standing at the same vertex, the game ends. Otherwise, Takahashi moves to a vertex of his choice that is adjacent to his current vertex.    * 2. If Takahashi and Aoki are standing at the same vertex, the game ends. Otherwise, Aoki moves to a vertex of his choice that is adjacent to his current vertex.    * 3. Go back to step 1.  Takahashi performs his moves so that the game ends as late as possible, while Aoki performs his moves so that the game ends as early as possible.  Find the number of moves Aoki will perform before the end of the game if both Takahashi and Aoki know each other's position and strategy.  It can be proved that the game is bound to end.

Submission attempts (in order):
n,u,v=(map(int,input().split()))
Graph=[]
for i in range(n):
    Graph.append([])
for i in range(n-1): 
    a,b=map(int,input().split())
    Graph[a-1].append(b-1)
    Graph[b-1].append(a-1)

def dfs(G,s,d): 
    mark[s]=1 
    depth[s]=d 
    for i in Graph[s]: 
        if mark[i]==0: 
            dfs(G,i,d+1) 

depth=[0]*n
mark=[0]*n
for i in range(n):
    dfs(Graph,u-1,0)
dT=depth

depth=[0]*n
mark=[0]*n
for i in range(n):
    dfs(Graph,v-1,0)
dA=depth

print(dT,dA)

dArrest=0
for i in range(n):
    if dA[i]>dT[i]>=dArrest: 
        dArrest=dA[i] 
print(dArrest-1)

import sys
sys.setrecursionlimit(1000000)

I=list(map(int,Input.split()))
n=I.pop(0)
u=I.pop(0)
v=I.pop(0)
Graph=[]
for i in range(n):
    Graph.append([])
for i in range(n-1): 
    a=I.pop(0)
    b=I.pop(0)
    Graph[a-1].append(b-1)
    Graph[b-1].append(a-1)

def dfs(s,b=-1): 
    for i in Graph[s]: 
        if i!=b: 
            depth[i]=depth[s]+1
            dfs(i,s) 

depth=[0]*n
dfs(u-1)
dT=depth

depth=[0]*n
dfs(v-1)
dA=depth

print(max(dT),max(dA))

dArrest=0
for i in range(n):
    if dA[i]>dT[i]: 
        dArrest=max(dArrest,dA[i]) 
print(dArrest-1)

import sys
sys.setrecursionlimit(1000000)

n,u,v=map(int,input().split())
Graph=[]
for i in range(n):
    Graph.append([])
for i in range(n-1): 
    a,b=map(int,input().split())
    Graph[a-1].append(b-1)
    Graph[b-1].append(a-1)

def dfs(s,b=-1): 
    for i in Graph[s]: 
        if i!=b: 
            depth[i]=depth[s]+1 
            dfs(i,s) 

depth=[0]*n
dfs(u-1)
dT=depth

depth=[0]*n
dfs(v-1)
dA=depth

dArrest=0
for i in range(n):
    if dA[i]>dT[i]: 
        dArrest=max(dArrest,dA[i]) 
print(dArrest-1)

Verdicts: Runtime Error

Runtime Error

Accepted
Problem: We have a tree with N vertices. The i-th edge connects Vertex A_i and B_i bidirectionally.  Takahashi is standing at Vertex u, and Aoki is standing at Vertex v.  Now, they will play a game of tag as follows:    * 1. If Takahashi and Aoki are standing at the same vertex, the game ends. Otherwise, Takahashi moves to a vertex of his choice that is adjacent to his current vertex.    * 2. If Takahashi and Aoki are standing at the same vertex, the game ends. Otherwise, Aoki moves to a vertex of his choice that is adjacent to his current vertex.    * 3. Go back to step 1.  Takahashi performs his moves so that the game ends as late as possible, while Aoki performs his moves so that the game ends as early as possible.  Find the number of moves Aoki will perform before the end of the game if both Takahashi and Aoki know each other's position and strategy.  It can be proved that the game is bound to end.

Submission attempts (in order):
from copy import deepcopy
N, u, v = map(int, input().split())
edge = [[] for _ in range(N)]
for _ in range(N-1):
    a, b = map(int, input().split())
    a, b = a-1, b-1
    edge[a].append(b)
    edge[b].append(a)

aoki = [0] * N
takahasi = [0] * N
dist_list = [0] * N
def dfs(pv, v, dist):
    dist_list[v] = dist
    for nv in edge[v]:
        if nv == pv:
            continue
        dfs(v, nv, dist+1)

dfs(-1,u-1,0)
takahasi = deepcopy(dist_list)
dist_list = [0] * N
dfs(-1,v-1,0)
aoki = deepcopy(dist_list)
print(takahasi)
print(aoki)



import sys
sys.setrecursionlimit(10**9)

N, u, v = map(int, input().split())
u, v = u-1, v-1
edge = [[] for _ in range(N)]
for i in range(N-1):
    a, b = map(int, input().split())
    a, b = a-1, b-1
    edge[a].append(b)
    edge[b].append(a)

taka = [0] * N
aoki = [0] * N

def dfs(v, pre, cost, i):
    for e in edge[v]:
        if e == pre:
            continue
        cost = dfs(e, v, cost, i+1)
    cost[v] = i
    return cost

taka = dfs(u, -1, [0] * N, 0)
aoki = dfs(v, -1, [0] * N, 0)

m = 0
for i in range(N):
    if taka[i] < aoki[i]:
        m = max(m, aoki[i])

print(m)


import sys
sys.setrecursionlimit(10**9)

N, u, v = map(int, input().split())
u, v = u-1, v-1
edge = [[] for _ in range(N)]
for i in range(N-1):
    a, b = map(int, input().split())
    a, b = a-1, b-1
    edge[a].append(b)
    edge[b].append(a)

taka = [0] * N
aoki = [0] * N

def dfs(v, pre, cost, i):
    for e in edge[v]:
        if e == pre:
            continue
        cost = dfs(e, v, cost, i+1)
    cost[v] = i
    return cost

taka = dfs(u, -1, [0] * N, 0)
aoki = dfs(v, -1, [0] * N, 0)

m = 0
for i in range(N):
    if taka[i] < aoki[i]:
        m = max(m, aoki[i])

print(m-1)


Verdicts: Runtime Error

Wrong Answer

Accepted
Problem: We have a tree with N vertices. The i-th edge connects Vertex A_i and B_i bidirectionally.  Takahashi is standing at Vertex u, and Aoki is standing at Vertex v.  Now, they will play a game of tag as follows:    * 1. If Takahashi and Aoki are standing at the same vertex, the game ends. Otherwise, Takahashi moves to a vertex of his choice that is adjacent to his current vertex.    * 2. If Takahashi and Aoki are standing at the same vertex, the game ends. Otherwise, Aoki moves to a vertex of his choice that is adjacent to his current vertex.    * 3. Go back to step 1.  Takahashi performs his moves so that the game ends as late as possible, while Aoki performs his moves so that the game ends as early as possible.  Find the number of moves Aoki will perform before the end of the game if both Takahashi and Aoki know each other's position and strategy.  It can be proved that the game is bound to end.

Submission attempts (in order):
import sys
sys.setrecursionlimit(10**7)
n,u,v = map(int,input().split())
ki = [[]for i in range(n)]
distt = [0]*n
dista = [0]*n
for i in range(n-1):
    a,b = map(int,input().split())
    a -= 1
    b -= 1
    ki[a].append(b)
    ki[b].append(a)
def dfst(now,last = -1):
    for next in ki[now]:
        if next == last:continue
        distt[next] = distt[now]+1
        dfst(next,now)
def dfsa(now,last = -1):
    for next in ki[now]:
        if next == last:continue
        dista[next] = dista[now]+1
        dfsa(next,now)
    
dfst(u)
dfsa(v)
ans = 0
for i in range(n):
    if dista[i]<=distt[i]:continue
    ans = max(ans,dista[i]-1)
print(ans)
    

import sys

sys.setrecursionlimit(10**7)
n,u,v = map(int,input().split())
u-=1
v-=1
ki = [[]for i in range(n)]
distt = [0]*n
dista = [0]*n
for i in range(n-1):
    a,b = map(int,input().split())
    a -= 1
    b -= 1
    ki[a].append(b)
    ki[b].append(a)
def dfst(now,last = -1):
    for next in ki[now]:
        if next == last:continue
        distt[next] = distt[now]+1
        dfst(next,now)
def dfsa(now,last = -1):
    for next in ki[now]:
        if next == last:continue
        dista[next] = dista[now]+1
        dfsa(next,now)
    
dfst(u)
dfsa(v)
ans = 0
for i in range(n):
    if dista[i]<=distt[i]:continue
    ans = max(ans,dista[i]-1)
print(ans)
    

Verdicts: Runtime Error

Accepted
Problem: We have a tree with N vertices. The i-th edge connects Vertex A_i and B_i bidirectionally.  Takahashi is standing at Vertex u, and Aoki is standing at Vertex v.  Now, they will play a game of tag as follows:    * 1. If Takahashi and Aoki are standing at the same vertex, the game ends. Otherwise, Takahashi moves to a vertex of his choice that is adjacent to his current vertex.    * 2. If Takahashi and Aoki are standing at the same vertex, the game ends. Otherwise, Aoki moves to a vertex of his choice that is adjacent to his current vertex.    * 3. Go back to step 1.  Takahashi performs his moves so that the game ends as late as possible, while Aoki performs his moves so that the game ends as early as possible.  Find the number of moves Aoki will perform before the end of the game if both Takahashi and Aoki know each other's position and strategy.  It can be proved that the game is bound to end.

Submission attempts (in order):
import sys
input = sys.stdin.readline


sys.setrecursionlimit(10**6)




n,u,v = (int(x) for x in input().split())

u,v = u-1,v-1

a = [[] for i in range(n)]

for i in range(n-1):
	
	l,m = (int(x)-1 for x in input().split())
	a[l].append(m)
	a[m].append(l)


dist_u = [0]*n
dist_v = [0]*n

# G:graph, cr:current node, seen:is seen?, dist:out;each distance
def dfs_u(cr,seen):
	seen[cr] = 1
	for i in G[cr]: 
		if seen[i] == 0: 
			dist_u[i] = dist_u[cr]+1
			dfs_u(i,seen)
# G:graph, cr:current node, seen:is seen?, dist:out;each distance
def dfs_v(cr,seen):
	seen[cr] = 1
	for i in G[cr]: 
		if seen[i] == 0: 
			dist_v[i] = dist_v[cr]+1
			dfs_v(i,seen)

dfs_u(u,[0]*n)

dfs_v(v,[0]*n)

for i in range(n):
	if dist_u[i] >= dist_v[i]: 
		dist_u[i] = -1

print(dist_v[dist_u.index(max(dist_u))] - 1)

import sys
input = sys.stdin.readline


sys.setrecursionlimit(10**6)

# G:graph, cr:current node, seen:is seen?, dist:out;each distance
def dfs(G,cr,seen,dist):
	seen[cr] = 1
	for i in G[cr]: 
		if seen[i] == 0: 
			dist[i] = dist[cr]+1
			dfs(G,i,seen,dist)


n,u,v = (int(x) for x in input().split())

u,v = u-1,v-1

a = [[] for i in range(n)]

for i in range(n-1):
	
	l,m = (int(x)-1 for x in input().split())
	a[l].append(m)
	a[m].append(l)


dist_u = [0]*n
dist_v = [0]*n

dfs(a,u,[0]*n,dist_u)

dfs(a,v,[0]*n,dist_v)

for i in range(n):
	if dist_u[i] >= dist_v[i]: 
		dist_v[i] = -1

print(max(dist_v) - 1)

Verdicts: Runtime Error

Accepted
Problem: We have a tree with N vertices. The i-th edge connects Vertex A_i and B_i bidirectionally.  Takahashi is standing at Vertex u, and Aoki is standing at Vertex v.  Now, they will play a game of tag as follows:    * 1. If Takahashi and Aoki are standing at the same vertex, the game ends. Otherwise, Takahashi moves to a vertex of his choice that is adjacent to his current vertex.    * 2. If Takahashi and Aoki are standing at the same vertex, the game ends. Otherwise, Aoki moves to a vertex of his choice that is adjacent to his current vertex.    * 3. Go back to step 1.  Takahashi performs his moves so that the game ends as late as possible, while Aoki performs his moves so that the game ends as early as possible.  Find the number of moves Aoki will perform before the end of the game if both Takahashi and Aoki know each other's position and strategy.  It can be proved that the game is bound to end.

Submission attempts (in order):
from collections import deque 
import sys
input = sys.stdin.readline  

N,u,v = map(int,input().split())
u -= 1; v -= 1
tree = [[] for _ in range(N)]
for i in range(N-1):
    A,B = map(int,input().split())
    tree[A-1].append(B-1)
    tree[B-1].append(A-1)

def bfs(start):
    q = deque()
    q.append([start,0])
    visited = set()
    lis = [0]*N
    while q:
        now,dis = q.popleft()
        if now in visited:
            continue
        lis[now] = dis
        visited.add(now)
        for node in tree[now]:
            q.append([node,dis+1])    
    return lis
ut = bfs(u)
vt = bfs(v)

q = deque()
q.append(0)
visited = set()
ans = 0
print(ut)
print(vt)
while q:
    now = q.popleft()
    if now in visited:
        continue
    elif ut[now] == vt[now]:
        ans = max(ans,vt[now])
        continue
    visited.add(now)
    if len(tree[now]) == 1:
        matsu = now
        kaisu = vt[matsu]-1
        ans = max(ans,kaisu)
    for child in tree[now]:
        q.append(child)
print(ans)


from collections import deque 
import sys
input = sys.stdin.readline  

N,u,v = map(int,input().split())
u -= 1; v -= 1
tree = [[] for _ in range(N)]
for i in range(N-1):
    A,B = map(int,input().split())
    tree[A-1].append(B-1)
    tree[B-1].append(A-1)

def bfs(start):
    q = deque()
    q.append([start,0])
    visited = set()
    lis = [0]*N
    while q:
        now,dis = q.popleft()
        if now in visited:
            continue
        lis[now] = dis
        visited.add(now)
        for node in tree[now]:
            q.append([node,dis+1])    
    return lis
ut = bfs(u)
vt = bfs(v)

ans = 0
for i in range(N):
    if ut[i] <vt[i]:
        ans = max(ans,vt[i])
print(ans-1)


Verdicts: Wrong Answer

Accepted
Problem: We have a tree with N vertices. The i-th edge connects Vertex A_i and B_i bidirectionally.  Takahashi is standing at Vertex u, and Aoki is standing at Vertex v.  Now, they will play a game of tag as follows:    * 1. If Takahashi and Aoki are standing at the same vertex, the game ends. Otherwise, Takahashi moves to a vertex of his choice that is adjacent to his current vertex.    * 2. If Takahashi and Aoki are standing at the same vertex, the game ends. Otherwise, Aoki moves to a vertex of his choice that is adjacent to his current vertex.    * 3. Go back to step 1.  Takahashi performs his moves so that the game ends as late as possible, while Aoki performs his moves so that the game ends as early as possible.  Find the number of moves Aoki will perform before the end of the game if both Takahashi and Aoki know each other's position and strategy.  It can be proved that the game is bound to end.

Submission attempts (in order):
#%%
import numpy as np
n,nige,oni = map(int,input().split())
aa = np.zeros((n,n),dtype=bool)
for i in range(n-1):
    a,b = map(int,input().split())
    aa[a-1][b-1] = 1
    aa[b-1][a-1] = 1
nigel = np.zeros(n)
nigel[nige-1] = 1
onil = np.zeros(n)
onil[oni-1] = 1
ans = 0
for i in range(n-1):
    nigel = np.any(aa[nigel] == True,axis = 0)
    if np.logical_and(onil,nigel) == nigel:
        print(ans)
        exit()
    ans += 1
    onil = np.any(aa[onil] == True,axis = 0)
    if np.logical_and(onil,nigel) == nigel:
        print(ans)
        exit()

#%%
import numpy as np
n,nige,oni = map(int,input().split())
aa = np.zeros((n,n),dtype=bool)
for i in range(n-1):
    a,b = map(int,input().split())
    aa[a-1][b-1] = 1
    aa[b-1][a-1] = 1
nigel = np.zeros(n)
nigel[nige-1] = 1
onil = np.zeros(n)
onil[oni-1] = 1
ans = 0
for i in range(n-1):
    nigel = np.any(aa[nigel] == True,axis = 0)
    if np.logical_and(onil,nigel) == nigel:
        print(ans)
        return
    ans += 1
    onil = np.any(aa[onil] == True,axis = 0)
    if np.logical_and(onil,nigel) == nigel:
        print(ans)
        return



ii = lambda : int(input())
mi = lambda : map(int,input().split())
li = lambda : list(map(int,input().split()))

n,u,v = mi()
data = [[] for _ in range(n)]
for i in range(n-1):
    a,b = mi()
    a -= 1
    b -= 1
    data[a].append(b)
    data[b].append(a)
inf = -1
oni = [inf]*n
ko = [inf]*n

from collections import deque

def start(hazime,alis):
    d = deque()
    d.append((hazime-1,0))
    alis[hazime-1] = 0
    while d:
        ima,num = d.popleft()
        num += 1
        for i in data[ima]:
            if alis[i] != inf:
                continue
            alis[i] = num 
            d.append((i,num))
start(u,ko)
start(v,oni)

if len(data[u]) == 1 and data[u] == v:
    print(0) 
    exit()
ans = -1
for i in range(n):
    if ko[i] <= oni[i]:
        ans = max(ans,oni[i]-1)
print(ans)



Verdicts: Runtime Error

Runtime Error

Accepted
Problem: We have a tree with N vertices. The i-th edge connects Vertex A_i and B_i bidirectionally.  Takahashi is standing at Vertex u, and Aoki is standing at Vertex v.  Now, they will play a game of tag as follows:    * 1. If Takahashi and Aoki are standing at the same vertex, the game ends. Otherwise, Takahashi moves to a vertex of his choice that is adjacent to his current vertex.    * 2. If Takahashi and Aoki are standing at the same vertex, the game ends. Otherwise, Aoki moves to a vertex of his choice that is adjacent to his current vertex.    * 3. Go back to step 1.  Takahashi performs his moves so that the game ends as late as possible, while Aoki performs his moves so that the game ends as early as possible.  Find the number of moves Aoki will perform before the end of the game if both Takahashi and Aoki know each other's position and strategy.  It can be proved that the game is bound to end.

Submission attempts (in order):
n,g,s = map(int,input().split())
G = [[] for _ in range(n)]
for i in range(n):
  a,b = map(int,input().split())
  G[a-1].append(b-1)
  G[b-1].append(a-1)
  
def dfs(V,v,d):
  if V[v] != -1:
    return
  V[v] = d
  for nv in G[v]:
    dfs(V,nv,d+1)
  
if len(G[g-1])==1 and G[g-1][0]==s-1:
  print(0)
else:
  vs = [-1]*n
  vg = [-1]*n
  dfs(vs,s-1,0)
  dfs(vg,g-1,0)
  
  ans = 0
  for i in range(n):
    if vs[i]>vg[i]:
      ans = max(ans,vs[i])
      
  print(ans-1)

import sys
sys.setrecursionlimit(10**6)

n,g,s = map(int,input().split())
G = [[] for _ in range(n)]
for i in range(n-1):
  a,b = map(int,input().split())
  G[a-1].append(b-1)
  G[b-1].append(a-1)
  
def dfs(V,v,d):
  if V[v] != -1:
    return
  V[v] = d
  for nv in G[v]:
    dfs(V,nv,d+1)
  
if len(G[g-1])==1 and G[g-1][0]==s-1:
  print(0)
else:
  vs = [-1]*n
  vg = [-1]*n
  dfs(vs,s-1,0)
  dfs(vg,g-1,0)
  
  ans = 0
  for i in range(n):
    if vs[i]>vg[i]:
      ans = max(ans,vs[i])
      
  print(ans-1)

Verdicts: Runtime Error

Accepted
Problem: Given are three integers A_1, A_2, and A_3.  If A_1+A_2+A_3 is greater than or equal to 22, print `bust`; otherwise, print `win`.

Submission attempts (in order):
list = list(map(int, input().split()))
if sum(list)<=22:
  print("bust")
else:
  print("win")

a, b, c = map(int, input().split())
list=[a, b, c]
if sum(list)>=22:
  print("bust")
else:
  print("win")

Verdicts: Wrong Answer

Accepted
Problem: Given are three integers A_1, A_2, and A_3.  If A_1+A_2+A_3 is greater than or equal to 22, print `bust`; otherwise, print `win`.

Submission attempts (in order):
A1, A2, A3 = map(int, input.split())
if A1 + A2 + A3 >= 22:
    print("bust")
else:
    print("win")

A1, A2, A3 = map(int, input().split())
if A1 + A2 + A3 >= 22:
    print("bust")
else:
    print("win")

Verdicts: Runtime Error

Accepted
Problem: Given are three integers A_1, A_2, and A_3.  If A_1+A_2+A_3 is greater than or equal to 22, print `bust`; otherwise, print `win`.

Submission attempts (in order):
print("win") if sum(map(int,input())) < 22 else print("bust")

print("win"if sum(map(int,input().split()))<22 else"bust")

Verdicts: Runtime Error

Accepted