D

D - Transit Tree Path

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Time limit : 2sec / Memory limit : 256MB

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≤10⁵
• 1≤a_i,b_i≤N(1≤i≤N−1)
• 1≤c_i≤10⁹(1≤i≤N−1)
• The given graph is a tree.
• 1≤Q≤10⁵
• 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)

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Input

Input is given from Standard Input in the following format:

N  
a1 b1 c1  
:  
aN−1 bN−1 cN−1
Q K
x1 y1
:  
xQ yQ

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.

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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

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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.

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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

linux有毒啊,昨天就写好了,一直说我编译不通过,找了好久都不知道怎么回事,早上就直接交一次,就过了,
什么情况啊,貌似他一直说我的v[x].push_back({y,z});这样写不能通过编译,哭死.
这个就是树啦,然后就用dfs就成了,代码挺短的,但是看到别人的怎么要差不多100行,可怕.

#include <bits/stdc++.h>
#define mem(a) memset(a,0,sizeof(a))
#define N 100010
#define ll long long int
using namespace std;
ll an[N];
struct Node{
  int to;
  ll cost;
};
vector<Node> v[N];
void dfs(int x,int fa){
  for(int i=0;i<v[x].size();i++){
    if(v[x][i].to==fa)
      continue;
    an[v[x][i].to]=an[x]+v[x][i].cost;
    dfs(v[x][i].to,x);
  }
}
int main(){
  int n;
  scanf("%d", &n);
  int x,y;
  ll z;
  for(int i=1;i<n;i++){
    scanf("%d%d%lld",&x,&y,&z);
    v[x].push_back({y,z});
    v[y].push_back({x,z});
  }
  int q,t;
  scanf("%d%d",&q,&t);
  dfs(t,-1);
  while(q--){
    scanf("%d%d",&x,&y);
    printf("%lld
",an[x]+an[y]);
  }
  return 0;
}