HDU 256 How far away ？

题目：

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

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <queue>
#include <algorithm>
#include <stack>
#include <vector>

using namespace std;
/*
 利用邻接链表储存数据
 通过深搜来找到每一条可以到达的路径，比较出来最短的一条
    如果用邻接矩阵会爆内存
 */
struct ac{
    int end,cost;
}www;

const int INF=0x3f3f3f3f;
int d[40004],v[40004],n,l=0,ans,w,yy,maxx;
vector<ac>q[40004];

void dfs(int x,int y){
    if(x==yy) {
        if(ans<maxx) maxx=ans;
        return ;
    }
    for(int i=0;i<q[x].size();i++){
        if(!v[q[x][i].end]){
            v[q[x][i].end]=1;
            ans+=q[x][i].cost;
            dfs(q[x][i].end,ans);
            ans-=q[x][i].cost;
            v[q[x][i].end]=0;
        }
    }
    
}

int main(){
    int a;
    cin>>a;
    while(a--){
        memset(q,0,sizeof(q));
        int b,aa,bb,cc,d1;
        cin>>n>>b;
        d1=n;
        while(--d1){
            cin>>aa>>bb>>cc;
            www.end=bb;
            www.cost=cc;
            q[aa].push_back(www);
            www.end=aa;
            q[bb].push_back(www);
        }
        
        int xx;
        while(b--){
            cin>>xx>>yy;
            memset(v,0,sizeof(v));
            ans=0;
            maxx=INF;
            dfs(xx,ans);
            cout<<maxx<<endl;
        }
        
    }
    
    return 0;
}