Problem:


Score : 500 points

Problem Statement

There are N towns numbered 1 to N and M roads. The i-th road connects Town A_i and Town B_i bidirectionally and has a length of C_i.

Takahashi will travel between these towns by car, passing through these roads. The fuel tank of his car can contain at most L liters of fuel, and one liter of fuel is consumed for each unit distance traveled. When visiting a town while traveling, he can full the tank (or choose not to do so). Travel that results in the tank becoming empty halfway on the road cannot be done.

Process the following Q queries:

The tank is now full. Find the minimum number of times he needs to full his tank while traveling from Town s_i to Town t_i. If Town t_i is unreachable, print -1.

Constraints

All values in input are integers.

2 \leq N \leq 300

0 \leq M \leq \frac{N(N-1)}{2}

1 \leq L \leq 10^9

1 \leq A_i, B_i \leq N

A_i \neq B_i

\left(A_i, B_i\right) \neq \left(A_j, B_j\right) (if i \neq j)

\left(A_i, B_i\right) \neq \left(B_j, A_j\right) (if i \neq j)

1 \leq C_i \leq 10^9

1 \leq Q \leq N\left(N-1\right)

1 \leq s_i, t_i \leq N

s_i \neq t_i

\left(s_i, t_i\right) \neq \left(s_j, t_j\right) (if i \neq j)

Input

Input is given from Standard Input in the following format:

N M L
A_1 B_1 C_1
:
A_M B_M C_M
Q
s_1 t_1
:
s_Q t_Q

Output

Print Q lines.

The i-th line should contain the minimum number of times the tank needs to be fulled while traveling from Town s_i to Town t_i. If Town t_i is unreachable, the line should contain -1 instead.

Sample Input 1

3 2 5
1 2 3
2 3 3
2
3 2
1 3

Sample Output 1

0
1

To travel from Town 3 to Town 2, we can use the second road to reach Town 2 without fueling the tank on the way.

To travel from Town 1 to Town 3, we can first use the first road to get to Town 2, full the tank, and use the second road to reach Town 3.

Sample Input 2

4 0 1
1
2 1

Sample Output 2

-1

There may be no road at all.

Sample Input 3

5 4 4
1 2 2
2 3 2
3 4 3
4 5 2
20
2 1
3 1
4 1
5 1
1 2
3 2
4 2
5 2
1 3
2 3
4 3
5 3
1 4
2 4
3 4
5 4
1 5
2 5
3 5
4 5

Sample Output 3

0
0
1
2
0
0
1
2
0
0
0
1
1
1
0
0
2
2
1
0



Sample input:
3 2 5
1 2 3
2 3 3
2
3 2
1 3


Sample output:
0
1

Submission status: Wrong Answer

Solution code:
#import sys
from collections import deque
printn = lambda x: sys.stdout.write(x)
inn = lambda : int(input())
inl   = lambda: list(map(int, input().split()))
inm   = lambda:      map(int, input().split()) 
DBG = True  and False
BIG = 1234567890

def ddprint(x):
  if DBG:
    print(x)

n,m,l = inm()

distance = [ [BIG] * n for i in range(n) ]
hops = [ [(BIG,0)] * n for i in range(n) ]
for i in range(n):
  distance[i][i] = 0
  hops[i][i] = (0,0)

dst = [ [] for i in range(n) ]
for i in range(m):
  a,b,c = inm()
  if c>l:
    continue
  a -= 1
  b -= 1
  distance[a][b] = distance[b][a] = c
  dst[a].append(b)
  dst[b].append(a)

for i in range(n):
  for j in range(n):
    if j==i:
      continue
    if hops[i][j] > hops[j][i]:
      hops[i][j] = hops[j][i]
  q = deque([ (x,(0,distance[i][x])) for x in dst[i] ])
  while len(q)>0:
    y,e = q.popleft()
    #ddprint("y {} e {} {}".format(y,e[0],e[1]))
    if e>=hops[i][y]:
      continue
    #ddprint("upd i {} y {}".format(i,y))
    hops[i][y] = e
    for f in dst[y]:
      ds = distance[y][f]
      #ddprint("y {} f {} ds {}".format(y,f,ds))
      if e[1]+ds > l:
        newd = (e[0]+1, ds)
      else:
        newd = (e[0], e[1]+ds)
      if newd < hops[i][f]:
        q.append((f,newd))

if DBG:
  for i in range(n):
    print(hops[i])

B2 = BIG//2
#for i in range(n):
#  for j in range(n):
#    if hops[i][j][0] > B2:
#      hops[i][j] = (-1,0)

p = inn()
for i in range(p):
  s,t = inm()
  #ddprint("q {} s {} t {}".format(i,s,t))
  x = hops[s-1][t-1][0]
  print(x if x<B2 else -1)

Submission status: Wrong Answer

Solution code:
N,M,L = map(int,input().split())
abc = [[int(i) for i in input().split()] for j in range(M)]
Q = int(input())
INF = 1<<30
Path = [[INF for i in range(N)] for j in range(N)]
AnsPath = [[INF for i in range(N)] for j in range(N)]

def PathInit():
    for i in range(N):
        Path[i][i] = 0

def wasialFloid(Path):
    for k in range(N):
        for i in range(N):
            for j in range(N):
                Path[i][j] = min(Path[i][j],Path[i][k]+Path[k][j])

PathInit()
for i,j,k in abc:
    Path[i-1][j-1] = k
    "双方向"

    Path[j-1][i-1] = k
wasialFloid(Path)
for i in range(N):
    for j in range(N):
        if Path[i][j] <= L:
            AnsPath[i][j] = 1
wasialFloid(AnsPath)
print("check")
for i in range(Q):
    s,t = map(int,input().split())
    if AnsPath[s-1][t-1] == INF:
        print(-1)
    else:
        print(AnsPath[s-1][t-1]-1)

Submission status: Wrong Answer

Solution code:
#O(ElogV)
import heapq
def dijkstra_heap(s,g):
    used = [True] * n #True:未確定
    used[s] = False
    edgelist = []
    for e in edge[s]:
        if e[0]<=l:
            heapq.heappush(edgelist,[[0,l-e[0]],e[1]])
    while len(edgelist):
        minedge = heapq.heappop(edgelist)
        #まだ使われてない頂点の中から最小の距離のものを探す
        if not used[minedge[1]]:
            continue
        v = minedge[1]
        if v==g:
            return minedge[0][0]
        nl = minedge[0][1]
        kai=minedge[0][0]
        used[v] = False
        for e in edge[v]:
            if used[e[1]]:
                if e[0]<=l and e[0]<=nl:
                    heapq.heappush(edgelist,[[kai,nl-e[0]],e[1]])
                elif e[0]<=l:
                    heapq.heappush(edgelist,[[kai+1,l-e[0]],e[1]])
                    
    return -1

################################
n,w,l = map(int,input().split()) #n:頂点数　w:辺の数

edge = [[] for i in range(n)]
#edge[i] : iから出る道の[重み,行先]の配列
for i in range(w):
    x,y,z = map(int,input().split())
    edge[x-1].append([z,y-1])
    edge[y-1].append([z,x-1])
q=int(input())
for i in range(q):
    sa,ga=map(int,input().split())
    print(dijkstra_heap(sa-1,ga-1))

Submission status: Wrong Answer

Solution code:
import numpy as np
N,M,L = map(int,input().split( ))
R = [[] for x in range(N+1)]
ST = []
def depth_search(start,end,gas,gasup):
    stack.append(start)
    for i in range(len(R[start])):
        if R[start][i][0] not in stack:
            now = R[start][i][0]
            if gas >= R[start][i][1]:
                gas = gas - R[start][i][1]
            else:
                gas = L - R[start][i][0]
                gasup+=1
            gasreco.append([gas,gasup])
            if now == end :
                global mingas
                if mingas == -1 or gasup < mingas:
                    mingas = gasup
                gasreco.pop()
                continue
            depth_search(now,end,gas,gasup)
    stack.pop()
    if stack == []:
        return mingas
    else:
        gasreco.pop()


for i in range(M):
    A, B, C = map(int,input().split( ))
    if C <= L:
        R[A].append([B,C])
        R[B].append([A,C])
Q = int(input())
for i in range(Q):
    ST.append(list(map(int,input().split( ))))
result = [-1 for _ in range(Q)]
for i in range(Q):
    gasreco = []
    stack = []
    mingas = -1
    mingasoline = depth_search(ST[i][0],ST[i][1],L,0)
    result[i] = mingasoline
for i in result:
    print(i)

Submission status: Wrong Answer

Solution code:
from scipy.sparse.csgraph import csgraph_from_dense
from scipy.sparse import csr_matrix
from scipy.sparse.csgraph import floyd_warshall
import numpy as np
N,M,L=map(int,input().split())
A=[0]*M
B=[0]*M
C=[0]*M
for i in range(M):
    A[i],B[i],C[i] = map(int,input().split())
    A[i] -= 1
    B[i] -= 1
Q=int(input())
ST=[list(map(int,input().split())) for _ in range(Q)]


G = csr_matrix((C, (A, B)), shape=(N, N))
d = floyd_warshall(G)
G=np.full((N,N), np.inf)
G[d<=L]=1
G = csr_matrix(G)
E = floyd_warshall(G)

for s,t in ST:
    if E[s-1][t-1]==float('inf'):
        print(-1)
    else:
        print(int(E[s-1][t-1]-1))

Submission status: Wrong Answer

Solution code:
from scipy.sparse.csgraph import floyd_warshall as fw

N, M, L = map(int, input().split())

S = [[float("inf") for i in range(N)] for i in range(N)]

for i in range(M):
    a, b, c = map(int, input().split())
    S[a-1][b-1] = c
    S[b-1][a-1] = c

Sdist = fw(S)

for i in range(N):
    for j in range(N):
        if Sdist[i][j]<=L:
            Sdist[i][j] = 1

Ssup = fw(Sdist)

Q = int(input())

for i in range(Q):
    s, t = map(int, input().split())
    if Ssup[s-1][t-1]==float("inf"):
        print(-1)
    else:
        print(int(Ssup[s-1][t-1]-1))

Submission status: Wrong Answer

Solution code:
def w(d):
    n=len(d)
    for k in range(n):
        for i in range(n):
            for j in range(n):
                d[i][j]=min(d[i][j],d[i][k]+d[k][j])
    return d
N,M,L = map(int, input().split())
inf=float('inf')
fuel=[[inf]*N for _ in range(N)]
for i in range(N):
    fuel[i][i]=0
for _ in range(M):
    a,b,c=map(int,input().split())
    fuel[a-1][b-1]=fuel[b-1][a-1]=c
Q=int(input())
ST=[list(map(int, input().split())) for _ in range(Q)]
fuel = w(fuel)
for i in range(N):
    for j in range(N):
        if 0 < fuel[i][j] <= L:
            fuel[i][j]=1
fuel = w(fuel)
for st in ST:
    print(fuel[st[0]-1][st[1]-1]-1 if fuel[st[0]-1][st[1]-1] != inf else -1)

Submission status: Wrong Answer

Solution code:
N,M,L = map(int, input().split())
inf=float('inf')
fuel=[[inf]*N for _ in range(N)]
for i in range(N):
    fuel[i][i]=0
for _ in range(M):
    a,b,c=map(int,input().split())
    fuel[a-1][b-1]=fuel[b-1][a-1]=c
Q=int(input())
ST=[list(map(int, input().split())) for _ in range(Q)]
for k in range(N):
    for i in range(N):
        for j in range(N):
            fuel[i][j]=min(fuel[i][j],fuel[i][k]+fuel[k][j])
for i in range(N):
    for j in range(N):
        if 0 < fuel[i][j] <= L:
            fuel[i][j]=1
for k in range(N):
    for i in range(N):
        for j in range(N):
            fuel[i][j]=min(fuel[i][j], fuel[i][k]+fuel[k][j])
for st in ST:
    print(fuel[st[0]-1][st[1]-1]-1 if fuel[st[0]-1][st[1]-1] != inf else -1)

Submission status: Wrong Answer

Solution code:
#O(ElogV)
import heapq
def dijkstra_heap(s,g):
    used = [True] * n #True:未確定
    used[s] = False
    edgelist = []
    for e in edge[s]:
        if e[0]<=l:
            heapq.heappush(edgelist,[[0,l-e[0]],e[1]])
    while len(edgelist):
        minedge = heapq.heappop(edgelist)
        #まだ使われてない頂点の中から最小の距離のものを探す
        if not used[minedge[1]]:
            continue
        v = minedge[1]
        if v==g:
            return minedge[0][0]
        nl = minedge[0][1]
        kai=minedge[0][0]
        used[v] = False
        for e in edge[v]:
            if used[e[1]]:
                if e[0]<=l and e[0]<=nl:
                    heapq.heappush(edgelist,[[kai,nl-e[0]],e[1]])
                elif e[0]<=l:
                    heapq.heappush(edgelist,[[kai+1,l],e[1]])
                    
    return -1

################################
n,w,l = map(int,input().split()) #n:頂点数　w:辺の数

edge = [[] for i in range(n)]
#edge[i] : iから出る道の[重み,行先]の配列
for i in range(w):
    x,y,z = map(int,input().split())
    edge[x-1].append([z,y-1])
    edge[y-1].append([z,x-1])
q=int(input())
for i in range(q):
    sa,ga=map(int,input().split())
    print(dijkstra_heap(sa-1,ga-1))

Submission status: Wrong Answer

Solution code:
import sys
input = sys.stdin.readline
N,M,L = map(int, input().split())
INF = float("inf")
dist1 = [[INF]*N for i in range(N)]
dist2 = [[INF]*N for i in range(N)]

for i in range(M):
  A,B,C = map(int, input().split())
  A -= 1
  B -= 1
  dist1[A][B] = C
  dist1[B][A] = C

for i in range(N):
  dist1[i][i] = 0
  dist2[i][i] = 0
  
for k in range(N):
  for i in range(N):
    for j in range(N):
      d = dist1[i][k]+dist1[k][j]
      if dist1[i][j] <= d:
        continue
      dist1[i][j] = d
for i in range(N):
  for j in range(N):
    if dist1[i][j] > L:
      continue
    dist2[i][j] = 1
for i in range(N):
  dist2[i][j] = 0
  
for k in range(N):
  for i in range(N):
    for j in range(N):
      d = dist2[i][k]+dist2[k][j]
      if dist2[i][j] < d:
        continue
      dist2[i][j] = d
Q = int(input())
for i in range(Q):
  s,t = map(int, input().split())
  print(dist2[s-1][t-1]-1 if dist2[s-1][t-1]<=L else -1)