How far away ？

Problem Description

There are n houses in the village and some bidirectional roads connecting them. Every day peole always like to ask like this "How far is it if I want to go from house A to house B"? Usually it hard to answer. But luckily int this village the answer is always unique, since the roads are built in the way that there is a unique simple path("simple" means you can't visit a place twice) between every two houses. Yout task is to answer all these curious people.

 

Input

First line is a single integer T(T<=10), indicating the number of test cases.
  For each test case,in the first line there are two numbers n(2<=n<=40000) and m (1<=m<=200),the number of houses and the number of queries. The following n-1 lines each consisting three numbers i,j,k, separated bu a single space, meaning that there is a road connecting house i and house j,with length k(0<k<=40000).The houses are labeled from 1 to n.
  Next m lines each has distinct integers i and j, you areato answer the distance between house i and house j.

 

Output

For each test case,output m lines. Each line represents the answer of the query. Output a bland line after each test case.

 

Sample Input

2
3 2
1 2 10
3 1 15
1 2
2 3

2 2
1 2 100
1 2
2 1

 

Sample Output

10
25
100
100

题解：深搜做的，g++提交栈会爆，必须c++提交.

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>

using namespace std;

struct Node
{
	int pos;
	int dis;
};

vector<Node> vec[40004];
bool visited[40004];

void dfs(int u,int v,int res)
{
	int k;
	visited[u] = true;     //访问过了 
	for(int i = 0;i < vec[u].size();i++)
	{
		k = vec[u][i].pos;
		if(visited[k])       //已经访问 
		{
			continue;
		}
		if(v == k)       //找到目的地 
		{
			printf("%d\n",res + vec[u][i].dis);
			return;
		}
		else
		{
			dfs(k,v,res + vec[u][i].dis);
		}
	}
}

int main()
{
	int ncase;
	cin>>ncase;
	int n,m;
	while(ncase--)
	{
		scanf("%d%d",&n,&m);
		Node p,q;
		int u,v,d;
		for(int i = 1;i < n;i++)
		{
			scanf("%d%d%d",&p.pos,&q.pos,&d);
			p.dis = q.dis = d;
			vec[p.pos].push_back(q);
			vec[q.pos].push_back(p);
		}
		
		for(int i = 0;i < m;i++)
		{
			scanf("%d%d",&u,&v);
			memset(visited,false,sizeof(visited));
			dfs(u,v,0);
		}
		
		for(int i = 1;i <= n;i++)
		{
			vec[i].clear();
		}
	}
	
	return 0;
}