Problem:


Score : 400 points

Problem Statement

You are given a tree with N vertices.

Here, a tree is a kind of graph, and more specifically, a connected undirected graph with N-1 edges, where N is the number of its vertices.

The i-th edge (1≤i≤N-1) connects Vertices a_i and b_i, and has a length of c_i.

You are also given Q queries and an integer K. In the j-th query (1≤j≤Q):

find the length of the shortest path from Vertex x_j and Vertex y_j via Vertex K.

Constraints

3≤N≤10^5 

1≤a_i,b_i≤N (1≤i≤N-1) 

1≤c_i≤10^9 (1≤i≤N-1) 

The given graph is a tree.

1≤Q≤10^5 

1≤K≤N 

1≤x_j,y_j≤N (1≤j≤Q) 

x_j≠y_j (1≤j≤Q) 

x_j≠K,y_j≠K (1≤j≤Q) 

Input

Input is given from Standard Input in the following format:  

N  
a_1 b_1 c_1  
:  
a_{N-1} b_{N-1} c_{N-1}
Q K
x_1 y_1
:  
x_{Q} y_{Q}

Output

Print the responses to the queries in Q lines.

In the j-th line j(1≤j≤Q), print the response to the j-th query.

Sample Input 1

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

Sample Output 1

3
2
4

The shortest paths for the three queries are as follows:

Query 1: Vertex 2 → Vertex 1 → Vertex 2 → Vertex 4 : Length 1+1+1=3 

Query 2: Vertex 2 → Vertex 1 → Vertex 3 : Length 1+1=2 

Query 3: Vertex 4 → Vertex 2 → Vertex 1 → Vertex 3 → Vertex 5 : Length 1+1+1+1=4 

Sample Input 2

7
1 2 1
1 3 3
1 4 5
1 5 7
1 6 9
1 7 11
3 2
1 3
4 5
6 7

Sample Output 2

5
14
22

The path for each query must pass Vertex K=2.

Sample Input 3

10
1 2 1000000000
2 3 1000000000
3 4 1000000000
4 5 1000000000
5 6 1000000000
6 7 1000000000
7 8 1000000000
8 9 1000000000
9 10 1000000000
1 1
9 10

Sample Output 3

17000000000



Sample input:
5
1 2 1
1 3 1
2 4 1
3 5 1
3 1
2 4
2 3
4 5


Sample output:
3
2
4

Submission status: Time Limit Exceeded

Solution code:
import sys
import heapq
from itertools import permutations
from bisect import bisect_left, bisect_right
from collections import Counter, deque
from fractions import gcd
from math import factorial, sqrt
INF = 1 << 60
sys.setrecursionlimit(10 ** 7)
input = sys.stdin.readline

class UnionFind():
    def __init__(self, n):
        self.n = n
        self.parents = [-1] * n

    def find(self, x):
        if self.parents[x] < 0:
            return x
        else:
            self.parents[x] = self.find(self.parents[x])
            return self.parents[x]

    def union(self, x, y):
        x = self.find(x)
        y = self.find(y)

        if x == y:
            return

        if self.parents[x] > self.parents[y]:
            x, y = y, x

        self.parents[x] += self.parents[y]
        self.parents[y] = x

    def size(self, x):
        return -self.parents[self.find(x)]

    def same(self, x, y):
        return self.find(x) == self.find(y)

    def members(self, x):
        root = self.find(x)
        return [i for i in range(self.n) if self.find(i) == root]

    def roots(self):
        return [i for i, x in enumerate(self.parents) if x < 0]

    def group_count(self):
        return len(self.roots())

    def all_group_members(self):
        return {r: self.members(r) for r in self.roots()}

    def __str__(self):
        return '\n'.join('{}: {}'.format(r, self.members(r)) for r in self.roots())

def dijkstra_heap(s,edge):
    #始点sから各頂点への最短距離
    d = [10**20] * n
    used = [True] * n #True:未確定
    d[s] = 0
    used[s] = False
    edgelist = []
    for a,b in edge[s]:
        heapq.heappush(edgelist,a*(10**6)+b)
    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%(10**6)]:
            continue
        v = minedge%(10**6)
        d[v] = minedge//(10**6)
        used[v] = False
        for e in edge[v]:
            if used[e[1]]:
                heapq.heappush(edgelist,(e[0]+d[v])*(10**6)+e[1])
    return d


def factorization(n):
    arr = []
    temp = n
    for i in range(2, int(-(-n**0.5//1))+1):
        if temp%i==0:
            cnt=0
            while temp%i==0:
                cnt+=1
                temp //= i
            arr.append([i, cnt])

    if temp!=1:
        arr.append([temp, 1])

    if arr==[]:
        arr.append([n, 1])

    return arr

#ここから書き始める
# def dfs(current, goal, cost, visited):
#     if current == goal:
#         return cost
#     if len(tree[current]) == 0:
#         return 0 
#     for i in tree[current]:
#         # print("i =", i)
#         if visited[i[0]] == 0:
#             return dfs(i[0], goal, cost + i[1])

def bfs(queue, goal):
    while queue:
        node  = queue.popleft()
        # print("node =", node)
        current = node[0]
        new_cost = node[1]
        cost = node[2]
        cost += new_cost
        if current == goal:
            return cost
        for i in tree[current]:
            queue.append([i[0], i[1], cost])

n = int(input())
tree = [[] for i in range(n)]
for i in range(n - 1):
    a, b, c = map(int, input().split())
    a -= 1
    b -= 1
    tree[a].append([b, c])
    tree[b].append([a, c])
# print(tree)
q, k = map(int, input().split())
k -= 1
dist = [0] * n
# visited = [0] * n
for i in range(n):
    queue = deque([[k, 0, 0]])
    # print(queue)
    dist[i] = bfs(queue, i)
# print("dist =", dist)
ans = [0] * q
for i in range(q):
    x, y = map(int, input().split())
    x -= 1
    y -= 1
    ans[i] = dist[x] + dist[y]
for i in ans:
    print(i)

Submission status: Time Limit Exceeded

Solution code:
import sys
import heapq
from itertools import permutations
from bisect import bisect_left, bisect_right
from collections import Counter, deque
from fractions import gcd
from math import factorial, sqrt
INF = 1 << 60
sys.setrecursionlimit(10 ** 7)
input = sys.stdin.readline

class UnionFind():
    def __init__(self, n):
        self.n = n
        self.parents = [-1] * n

    def find(self, x):
        if self.parents[x] < 0:
            return x
        else:
            self.parents[x] = self.find(self.parents[x])
            return self.parents[x]

    def union(self, x, y):
        x = self.find(x)
        y = self.find(y)

        if x == y:
            return

        if self.parents[x] > self.parents[y]:
            x, y = y, x

        self.parents[x] += self.parents[y]
        self.parents[y] = x

    def size(self, x):
        return -self.parents[self.find(x)]

    def same(self, x, y):
        return self.find(x) == self.find(y)

    def members(self, x):
        root = self.find(x)
        return [i for i in range(self.n) if self.find(i) == root]

    def roots(self):
        return [i for i, x in enumerate(self.parents) if x < 0]

    def group_count(self):
        return len(self.roots())

    def all_group_members(self):
        return {r: self.members(r) for r in self.roots()}

    def __str__(self):
        return '\n'.join('{}: {}'.format(r, self.members(r)) for r in self.roots())

def dijkstra_heap(s,edge):
    #始点sから各頂点への最短距離
    d = [10**20] * n
    used = [True] * n #True:未確定
    d[s] = 0
    used[s] = False
    edgelist = []
    for a,b in edge[s]:
        heapq.heappush(edgelist,a*(10**6)+b)
    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%(10**6)]:
            continue
        v = minedge%(10**6)
        d[v] = minedge//(10**6)
        used[v] = False
        for e in edge[v]:
            if used[e[1]]:
                heapq.heappush(edgelist,(e[0]+d[v])*(10**6)+e[1])
    return d


def factorization(n):
    arr = []
    temp = n
    for i in range(2, int(-(-n**0.5//1))+1):
        if temp%i==0:
            cnt=0
            while temp%i==0:
                cnt+=1
                temp //= i
            arr.append([i, cnt])

    if temp!=1:
        arr.append([temp, 1])

    if arr==[]:
        arr.append([n, 1])

    return arr

#ここから書き始める
def dfs(current, goal, cost):
    if current == goal:
        return cost
    if len(tree[current]) == 0:
        return 0 
    for i in tree[current]:
        # print("i =", i)
        return dfs(i[0], goal, cost + i[1])

n = int(input())
tree = [[] for i in range(n)]
for i in range(n - 1):
    a, b, c = map(int, input().split())
    a -= 1
    b -= 1
    tree[a].append([b, c])
    tree[b].append([a, c])
# print(tree)
q, k = map(int, input().split())
k -= 1
dist = [0] * n
for i in range(n):
    dist[i] = dfs(i, k, 0)
# print("dist =", dist)
ans = [0] * q
for i in range(q):
    x, y = map(int, input().split())
    x -= 1
    y -= 1
    ans[i] = dist[x] + dist[y]
for i in ans:
    print(i)

Submission status: Time Limit Exceeded

Solution code:
import sys
import heapq
from itertools import permutations
from bisect import bisect_left, bisect_right
from collections import Counter, deque
from fractions import gcd
from math import factorial, sqrt
INF = 1 << 60
sys.setrecursionlimit(10 ** 6)

class UnionFind():
    def __init__(self, n):
        self.n = n
        self.parents = [-1] * n

    def find(self, x):
        if self.parents[x] < 0:
            return x
        else:
            self.parents[x] = self.find(self.parents[x])
            return self.parents[x]

    def union(self, x, y):
        x = self.find(x)
        y = self.find(y)

        if x == y:
            return

        if self.parents[x] > self.parents[y]:
            x, y = y, x

        self.parents[x] += self.parents[y]
        self.parents[y] = x

    def size(self, x):
        return -self.parents[self.find(x)]

    def same(self, x, y):
        return self.find(x) == self.find(y)

    def members(self, x):
        root = self.find(x)
        return [i for i in range(self.n) if self.find(i) == root]

    def roots(self):
        return [i for i, x in enumerate(self.parents) if x < 0]

    def group_count(self):
        return len(self.roots())

    def all_group_members(self):
        return {r: self.members(r) for r in self.roots()}

    def __str__(self):
        return '\n'.join('{}: {}'.format(r, self.members(r)) for r in self.roots())

def dijkstra_heap(s,edge):
    #始点sから各頂点への最短距離
    d = [10**20] * n
    used = [True] * n #True:未確定
    d[s] = 0
    used[s] = False
    edgelist = []
    for a,b in edge[s]:
        heapq.heappush(edgelist,a*(10**6)+b)
    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%(10**6)]:
            continue
        v = minedge%(10**6)
        d[v] = minedge//(10**6)
        used[v] = False
        for e in edge[v]:
            if used[e[1]]:
                heapq.heappush(edgelist,(e[0]+d[v])*(10**6)+e[1])
    return d


def factorization(n):
    arr = []
    temp = n
    for i in range(2, int(-(-n**0.5//1))+1):
        if temp%i==0:
            cnt=0
            while temp%i==0:
                cnt+=1
                temp //= i
            arr.append([i, cnt])

    if temp!=1:
        arr.append([temp, 1])

    if arr==[]:
        arr.append([n, 1])

    return arr

#ここから書き始める
input = sys.stdin.readline

n = int(input())
edge = [[] for i in range(n)]
for i in range(n - 1):
    a, b, c = map(int, input().split())
    a -= 1
    b -= 1
    edge[a].append((c, b))
    edge[b].append((c, a))
q, k = map(int, input().split())   
ans = [0] * q
dist = [0] * n
for i in range(n):
    dist[i] = dijkstra_heap(i, edge)[k - 1]
for i in range(q):
    x, y = map(int, input().split())
    # print(dist[x - 1] + dist[y - 1])
    ans[i] = dist[x - 1] + dist[y - 1]
for i in range(q):
    print(ans[i])

Submission status: Time Limit Exceeded

Solution code:
import sys
import heapq
from itertools import permutations
from bisect import bisect_left, bisect_right
from collections import Counter, deque
from fractions import gcd
from math import factorial, sqrt
INF = 1 << 60
sys.setrecursionlimit(10 ** 6)

class UnionFind():
    def __init__(self, n):
        self.n = n
        self.parents = [-1] * n

    def find(self, x):
        if self.parents[x] < 0:
            return x
        else:
            self.parents[x] = self.find(self.parents[x])
            return self.parents[x]

    def union(self, x, y):
        x = self.find(x)
        y = self.find(y)

        if x == y:
            return

        if self.parents[x] > self.parents[y]:
            x, y = y, x

        self.parents[x] += self.parents[y]
        self.parents[y] = x

    def size(self, x):
        return -self.parents[self.find(x)]

    def same(self, x, y):
        return self.find(x) == self.find(y)

    def members(self, x):
        root = self.find(x)
        return [i for i in range(self.n) if self.find(i) == root]

    def roots(self):
        return [i for i, x in enumerate(self.parents) if x < 0]

    def group_count(self):
        return len(self.roots())

    def all_group_members(self):
        return {r: self.members(r) for r in self.roots()}

    def __str__(self):
        return '\n'.join('{}: {}'.format(r, self.members(r)) for r in self.roots())

def dijkstra_heap(s,edge):
    #始点sから各頂点への最短距離
    d = [10**20] * n
    used = [True] * n #True:未確定
    d[s] = 0
    used[s] = False
    edgelist = []
    for a,b in edge[s]:
        heapq.heappush(edgelist,a*(10**6)+b)
    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%(10**6)]:
            continue
        v = minedge%(10**6)
        d[v] = minedge//(10**6)
        used[v] = False
        for e in edge[v]:
            if used[e[1]]:
                heapq.heappush(edgelist,(e[0]+d[v])*(10**6)+e[1])
    return d


def factorization(n):
    arr = []
    temp = n
    for i in range(2, int(-(-n**0.5//1))+1):
        if temp%i==0:
            cnt=0
            while temp%i==0:
                cnt+=1
                temp //= i
            arr.append([i, cnt])

    if temp!=1:
        arr.append([temp, 1])

    if arr==[]:
        arr.append([n, 1])

    return arr

#ここから書き始める
input = sys.stdin.readline

n = int(input())
edge = [[] for i in range(n)]
for i in range(n - 1):
    a, b, c = map(int, input().split())
    a -= 1
    b -= 1
    edge[a].append([c, b])
    edge[b].append([c, a])
q, k = map(int, input().split())   
ans = [0] * q
dist = [0] * n
for i in range(n):
    dist[i] = dijkstra_heap(i, edge)[k - 1]
for i in range(q):
    x, y = map(int, input().split())
    # print(dist[x - 1] + dist[y - 1])
    ans[i] = dist[x - 1] + dist[y - 1]
for i in range(q):
    print(ans[i])

Submission status: Time Limit Exceeded

Solution code:
import sys
import heapq
from itertools import permutations
from bisect import bisect_left, bisect_right
from collections import Counter, deque
from fractions import gcd
from math import factorial, sqrt
INF = 1 << 60
sys.setrecursionlimit(10 ** 6)

class UnionFind():
    def __init__(self, n):
        self.n = n
        self.parents = [-1] * n

    def find(self, x):
        if self.parents[x] < 0:
            return x
        else:
            self.parents[x] = self.find(self.parents[x])
            return self.parents[x]

    def union(self, x, y):
        x = self.find(x)
        y = self.find(y)

        if x == y:
            return

        if self.parents[x] > self.parents[y]:
            x, y = y, x

        self.parents[x] += self.parents[y]
        self.parents[y] = x

    def size(self, x):
        return -self.parents[self.find(x)]

    def same(self, x, y):
        return self.find(x) == self.find(y)

    def members(self, x):
        root = self.find(x)
        return [i for i in range(self.n) if self.find(i) == root]

    def roots(self):
        return [i for i, x in enumerate(self.parents) if x < 0]

    def group_count(self):
        return len(self.roots())

    def all_group_members(self):
        return {r: self.members(r) for r in self.roots()}

    def __str__(self):
        return '\n'.join('{}: {}'.format(r, self.members(r)) for r in self.roots())

def dijkstra_heap(s,edge):
    #始点sから各頂点への最短距離
    d = [10**20] * n
    used = [True] * n #True:未確定
    d[s] = 0
    used[s] = False
    edgelist = []
    for a,b in edge[s]:
        heapq.heappush(edgelist,a*(10**6)+b)
    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%(10**6)]:
            continue
        v = minedge%(10**6)
        d[v] = minedge//(10**6)
        used[v] = False
        for e in edge[v]:
            if used[e[1]]:
                heapq.heappush(edgelist,(e[0]+d[v])*(10**6)+e[1])
    return d


def factorization(n):
    arr = []
    temp = n
    for i in range(2, int(-(-n**0.5//1))+1):
        if temp%i==0:
            cnt=0
            while temp%i==0:
                cnt+=1
                temp //= i
            arr.append([i, cnt])

    if temp!=1:
        arr.append([temp, 1])

    if arr==[]:
        arr.append([n, 1])

    return arr

#ここから書き始める
n = int(input())
edge = [[] for i in range(n)]
for i in range(n - 1):
    a, b, c = map(int, input().split())
    a -= 1
    b -= 1
    edge[a].append([c, b])
    edge[b].append([c, a])
q, k = map(int, input().split())   
ans = [0] * q
dist = [0] * n
for i in range(n):
    dist[i] = dijkstra_heap(i, edge)[k - 1]
for i in range(q):
    x, y = map(int, input().split())
    # print(dist[x - 1] + dist[y - 1])
    ans[i] = dist[x - 1] + dist[y - 1]
for i in range(q):
    print(ans[i])

Submission status: Time Limit Exceeded

Solution code:
import sys
from heapq import heappush, heappop, heappushpop, heapify, heapreplace
from itertools import permutations
from bisect import bisect_left, bisect_right
from collections import Counter, deque
from fractions import gcd
from math import factorial, sqrt
INF = 1 << 60
sys.setrecursionlimit(10 ** 6)

class UnionFind():
    def __init__(self, n):
        self.n = n
        self.parents = [-1] * n

    def find(self, x):
        if self.parents[x] < 0:
            return x
        else:
            self.parents[x] = self.find(self.parents[x])
            return self.parents[x]

    def union(self, x, y):
        x = self.find(x)
        y = self.find(y)

        if x == y:
            return

        if self.parents[x] > self.parents[y]:
            x, y = y, x

        self.parents[x] += self.parents[y]
        self.parents[y] = x

    def size(self, x):
        return -self.parents[self.find(x)]

    def same(self, x, y):
        return self.find(x) == self.find(y)

    def members(self, x):
        root = self.find(x)
        return [i for i in range(self.n) if self.find(i) == root]

    def roots(self):
        return [i for i, x in enumerate(self.parents) if x < 0]

    def group_count(self):
        return len(self.roots())

    def all_group_members(self):
        return {r: self.members(r) for r in self.roots()}

    def __str__(self):
        return '\n'.join('{}: {}'.format(r, self.members(r)) for r in self.roots())

def dijkstra(s,n,w,cost):
    #始点sから各頂点への最短距離
    #n:頂点数,　w:辺の数, cost[u][v] : 辺uvのコスト(存在しないときはinf)
    d = [float("inf")] * n
    used = [False] * n
    d[s] = 0
    
    while True:
        v = -1
        #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
        for i in range(n):
            if (not used[i]) and (v == -1):
                v = i
            elif (not used[i]) and d[i] < d[v]:
                v = i
        if v == -1:
            break
        used[v] = True
            
        for j in range(n):
            d[j] = min(d[j],d[v]+cost[v][j])
    return d


def factorization(n):
    arr = []
    temp = n
    for i in range(2, int(-(-n**0.5//1))+1):
        if temp%i==0:
            cnt=0
            while temp%i==0:
                cnt+=1
                temp //= i
            arr.append([i, cnt])

    if temp!=1:
        arr.append([temp, 1])

    if arr==[]:
        arr.append([n, 1])

    return arr

#ここから書き始める
n = int(input())
cost = [[INF for i in range(n)] for i in range(n)] 
for i in range(n - 1):
    a, b, c = map(int, input().split())
    cost[a - 1][b - 1] = c
    cost[b - 1][a - 1] = c
q, k = map(int, input().split())   
ans = [0] * q
to_k = [0] * n
for i in range(n):
    to_k[i] = dijkstra(i, n, n - 1, cost)[k - 1]
for i in range(q):
    x, y = map(int, input().split())
    print(to_k[x - 1] + to_k[y - 1])

Submission status: Time Limit Exceeded

Solution code:
import sys
from heapq import heappush, heappop, heappushpop, heapify, heapreplace
from itertools import permutations
from bisect import bisect_left, bisect_right
from collections import Counter, deque
from fractions import gcd
from math import factorial, sqrt
INF = 1 << 60
sys.setrecursionlimit(10 ** 6)

class UnionFind():
    def __init__(self, n):
        self.n = n
        self.parents = [-1] * n

    def find(self, x):
        if self.parents[x] < 0:
            return x
        else:
            self.parents[x] = self.find(self.parents[x])
            return self.parents[x]

    def union(self, x, y):
        x = self.find(x)
        y = self.find(y)

        if x == y:
            return

        if self.parents[x] > self.parents[y]:
            x, y = y, x

        self.parents[x] += self.parents[y]
        self.parents[y] = x

    def size(self, x):
        return -self.parents[self.find(x)]

    def same(self, x, y):
        return self.find(x) == self.find(y)

    def members(self, x):
        root = self.find(x)
        return [i for i in range(self.n) if self.find(i) == root]

    def roots(self):
        return [i for i, x in enumerate(self.parents) if x < 0]

    def group_count(self):
        return len(self.roots())

    def all_group_members(self):
        return {r: self.members(r) for r in self.roots()}

    def __str__(self):
        return '\n'.join('{}: {}'.format(r, self.members(r)) for r in self.roots())

def dijkstra(s,n,w,cost):
    #始点sから各頂点への最短距離
    #n:頂点数,　w:辺の数, cost[u][v] : 辺uvのコスト(存在しないときはinf)
    d = [float("inf")] * n
    used = [False] * n
    d[s] = 0
    
    while True:
        v = -1
        #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
        for i in range(n):
            if (not used[i]) and (v == -1):
                v = i
            elif (not used[i]) and d[i] < d[v]:
                v = i
        if v == -1:
            break
        used[v] = True
            
        for j in range(n):
            d[j] = min(d[j],d[v]+cost[v][j])
    return d


def factorization(n):
    arr = []
    temp = n
    for i in range(2, int(-(-n**0.5//1))+1):
        if temp%i==0:
            cnt=0
            while temp%i==0:
                cnt+=1
                temp //= i
            arr.append([i, cnt])

    if temp!=1:
        arr.append([temp, 1])

    if arr==[]:
        arr.append([n, 1])

    return arr

#ここから書き始める
n = int(input())
cost = [[INF for i in range(n)] for i in range(n)] 
for i in range(n - 1):
    a, b, c = map(int, input().split())
    cost[a - 1][b - 1] = c
    cost[b - 1][a - 1] = c
q, k = map(int, input().split())   
ans = [0] * q
to_k = [0] * n
for i in range(n):
    to_k[i] = dijkstra(i, n, n - 1, cost)[k - 1]
for i in range(q):
    x, y = map(int, input().split())
    ans[i] = to_k[x - 1] + to_k[y - 1]
for i in ans:
    print(i)

Submission status: Time Limit Exceeded

Solution code:
import sys
from heapq import heappush, heappop, heappushpop, heapify, heapreplace
from itertools import permutations
from bisect import bisect_left, bisect_right
from collections import Counter, deque
from fractions import gcd
from math import factorial, sqrt
INF = 1 << 60
sys.setrecursionlimit(10 ** 6)

class UnionFind():
    def __init__(self, n):
        self.n = n
        self.parents = [-1] * n

    def find(self, x):
        if self.parents[x] < 0:
            return x
        else:
            self.parents[x] = self.find(self.parents[x])
            return self.parents[x]

    def union(self, x, y):
        x = self.find(x)
        y = self.find(y)

        if x == y:
            return

        if self.parents[x] > self.parents[y]:
            x, y = y, x

        self.parents[x] += self.parents[y]
        self.parents[y] = x

    def size(self, x):
        return -self.parents[self.find(x)]

    def same(self, x, y):
        return self.find(x) == self.find(y)

    def members(self, x):
        root = self.find(x)
        return [i for i in range(self.n) if self.find(i) == root]

    def roots(self):
        return [i for i, x in enumerate(self.parents) if x < 0]

    def group_count(self):
        return len(self.roots())

    def all_group_members(self):
        return {r: self.members(r) for r in self.roots()}

    def __str__(self):
        return '\n'.join('{}: {}'.format(r, self.members(r)) for r in self.roots())

def dijkstra(s,n,w,cost):
    #始点sから各頂点への最短距離
    #n:頂点数,　w:辺の数, cost[u][v] : 辺uvのコスト(存在しないときはinf)
    d = [float("inf")] * n
    used = [False] * n
    d[s] = 0
    
    while True:
        v = -1
        #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
        for i in range(n):
            if (not used[i]) and (v == -1):
                v = i
            elif (not used[i]) and d[i] < d[v]:
                v = i
        if v == -1:
            break
        used[v] = True
            
        for j in range(n):
            d[j] = min(d[j],d[v]+cost[v][j])
    return d


def factorization(n):
    arr = []
    temp = n
    for i in range(2, int(-(-n**0.5//1))+1):
        if temp%i==0:
            cnt=0
            while temp%i==0:
                cnt+=1
                temp //= i
            arr.append([i, cnt])

    if temp!=1:
        arr.append([temp, 1])

    if arr==[]:
        arr.append([n, 1])

    return arr

#ここから書き始める
n = int(input())
cost = [[INF for i in range(n)] for i in range(n)] 
for i in range(n - 1):
    a, b, c = map(int, input().split())
    cost[a - 1][b - 1] = c
    cost[b - 1][a - 1] = c
q, k = map(int, input().split())   
ans = [0] * q
to_k = [0] * n
to_y = [0] * n
for i in range(n):
    to_k[i] = dijkstra(i, n, n - 1, cost)[k - 1]
    to_y[i] = dijkstra(k - 1, n, n - 1, cost)[i]
for i in range(q):
    x, y = map(int, input().split())
    ans[i] = to_k[x - 1] + to_y[y - 1]
for i in ans:
    print(i)

Submission status: Time Limit Exceeded

Solution code:
import sys
from heapq import heappush, heappop, heappushpop, heapify, heapreplace
from itertools import permutations
from bisect import bisect_left, bisect_right
from collections import Counter, deque
from fractions import gcd
from math import factorial, sqrt
INF = 1 << 60
sys.setrecursionlimit(10 ** 6)

class UnionFind():
    def __init__(self, n):
        self.n = n
        self.parents = [-1] * n

    def find(self, x):
        if self.parents[x] < 0:
            return x
        else:
            self.parents[x] = self.find(self.parents[x])
            return self.parents[x]

    def union(self, x, y):
        x = self.find(x)
        y = self.find(y)

        if x == y:
            return

        if self.parents[x] > self.parents[y]:
            x, y = y, x

        self.parents[x] += self.parents[y]
        self.parents[y] = x

    def size(self, x):
        return -self.parents[self.find(x)]

    def same(self, x, y):
        return self.find(x) == self.find(y)

    def members(self, x):
        root = self.find(x)
        return [i for i in range(self.n) if self.find(i) == root]

    def roots(self):
        return [i for i, x in enumerate(self.parents) if x < 0]

    def group_count(self):
        return len(self.roots())

    def all_group_members(self):
        return {r: self.members(r) for r in self.roots()}

    def __str__(self):
        return '\n'.join('{}: {}'.format(r, self.members(r)) for r in self.roots())

def dijkstra(s,n,w,cost):
    #始点sから各頂点への最短距離
    #n:頂点数,　w:辺の数, cost[u][v] : 辺uvのコスト(存在しないときはinf)
    d = [float("inf")] * n
    used = [False] * n
    d[s] = 0
    
    while True:
        v = -1
        #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
        for i in range(n):
            if (not used[i]) and (v == -1):
               v = i
            elif (not used[i]) and d[i] < d[v]:
                v = i
        if v == -1:
               break
        used[v] = True
               
        for j in range(n):
               d[j] = min(d[j],d[v]+cost[v][j])
    return d


def factorization(n):
    arr = []
    temp = n
    for i in range(2, int(-(-n**0.5//1))+1):
        if temp%i==0:
            cnt=0
            while temp%i==0:
                cnt+=1
                temp //= i
            arr.append([i, cnt])

    if temp!=1:
        arr.append([temp, 1])

    if arr==[]:
        arr.append([n, 1])

    return arr

#ここから書き始める
n = int(input())
cost = [[INF for i in range(n)] for i in range(n)] 
for i in range(n - 1):
    a, b, c = map(int, input().split())
    cost[a - 1][b - 1] = c
    cost[b - 1][a - 1] = c
q, k = map(int, input().split())   
ans = [0] * q
to_k = [0] * n
to_y = [0] * n
for i in range(n):
    to_k[i] = dijkstra(i, n, n - 1, cost)[k - 1]
    to_y[i] = dijkstra(k - 1, n, n - 1, cost)[i]
for i in range(q):
    x, y = map(int, input().split())
    ans[i] = to_k[x - 1] + to_y[y - 1]
for i in ans:
    print(i)

Submission status: Time Limit Exceeded

Solution code:
import sys
from heapq import heappush, heappop, heappushpop, heapify, heapreplace
from itertools import permutations
from bisect import bisect_left, bisect_right
from collections import Counter, deque
from fractions import gcd
from math import factorial, sqrt
INF = 1 << 60
sys.setrecursionlimit(10 ** 6)

class UnionFind():
    def __init__(self, n):
        self.n = n
        self.parents = [-1] * n

    def find(self, x):
        if self.parents[x] < 0:
            return x
        else:
            self.parents[x] = self.find(self.parents[x])
            return self.parents[x]

    def union(self, x, y):
        x = self.find(x)
        y = self.find(y)

        if x == y:
            return

        if self.parents[x] > self.parents[y]:
            x, y = y, x

        self.parents[x] += self.parents[y]
        self.parents[y] = x

    def size(self, x):
        return -self.parents[self.find(x)]

    def same(self, x, y):
        return self.find(x) == self.find(y)

    def members(self, x):
        root = self.find(x)
        return [i for i in range(self.n) if self.find(i) == root]

    def roots(self):
        return [i for i, x in enumerate(self.parents) if x < 0]

    def group_count(self):
        return len(self.roots())

    def all_group_members(self):
        return {r: self.members(r) for r in self.roots()}

    def __str__(self):
        return '\n'.join('{}: {}'.format(r, self.members(r)) for r in self.roots())

def dijkstra(s,n,w,cost):
    #始点sから各頂点への最短距離
    #n:頂点数,　w:辺の数, cost[u][v] : 辺uvのコスト(存在しないときはinf)
    d = [float("inf")] * n
    used = [False] * n
    d[s] = 0
    
    while True:
        v = -1
        #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
        for i in range(n):
            if (not used[i]) and (v == -1):
               v = i
            elif (not used[i]) and d[i] < d[v]:
                v = i
        if v == -1:
               break
        used[v] = True
               
        for j in range(n):
               d[j] = min(d[j],d[v]+cost[v][j])
    return d


def factorization(n):
    arr = []
    temp = n
    for i in range(2, int(-(-n**0.5//1))+1):
        if temp%i==0:
            cnt=0
            while temp%i==0:
                cnt+=1
                temp //= i
            arr.append([i, cnt])

    if temp!=1:
        arr.append([temp, 1])

    if arr==[]:
        arr.append([n, 1])

    return arr

#ここから書き始める
n = int(input())
cost = [[INF for i in range(n)] for i in range(n)] 
for i in range(n - 1):
    a, b, c = map(int, input().split())
    cost[a - 1][b - 1] = c
    cost[b - 1][a - 1] = c
q, k = map(int, input().split())   
ans = [0] * q
for i in range(q):
    x, y = map(int, input().split())
    to_k = dijkstra(x - 1, n, n - 1, cost)[k - 1]
    to_y = dijkstra(k - 1, n, n - 1, cost)[y - 1]
    ans[i] = to_k + to_y
    # print(ans)
for i in ans:
    print(i)