How far away ？(DFS)

How far away ？

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5766    Accepted Submission(s): 2166

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
   
   
   
   

 

Source

ECJTU 2009 Spring Contest

AC代码：

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

using namespace std;

const int M = 4 * 1e4 + 100;
typedef long long ll;
typedef pair<int,int> P;
vector<P>G[M];
int vis[M],isGO,e;

void dfs(int x, int cost)
{
    if(isGO || vis[x]) return ;
    if(x == e)//数据太大，要改成全局变量
    {
        isGO = 1;
        printf("%d\n",cost);
        return ;
    }
    vis[x] = 1;
    for(int i = 0; i < G[x].size(); i++)
    {
        P a = G[x][i];
        dfs(a.first,cost + a.second);
    }
    vis[x] = 0;
}

void solve()
{
    int x,y,c,n,m;
    scanf("%d %d",&n,&m);
    for(int i = 1; i < n; i++)
    {
        scanf("%d %d %d",&x,&y,&c);
        G[x].push_back(make_pair(y,c));
        G[y].push_back(make_pair(x,c));
    }
    while(m--)
    {
        isGO = 0;
        scanf("%d %d",&x,&e);
        dfs(x,0);
    }
    for(int i = 1; i <= n; i++) G[i].clear();
}

int main()
{
    int T,cnt = 0;
    scanf("%d",&T);
    while(T--)
    {
        solve();
    }
    return 0;
}