How far away ？（dfs水题）

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

dfs+vector就可以过，vector真是个好东西。总感觉这个题目出水了，40000的空间肯定不能开个二维空间，vector真的可以大大的减少空间。
代码如下：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#define ll long long
using namespace std;

const int maxx=4e4+10;
struct node{
	int x;
	int w;
}p;
vector<struct node>pp[maxx];//vector里面存结构体
int vis[maxx];
int n,m;
int flag;

void dfs(int u,int ans,int v)
{
	if(vis[u]) return ;
	if(flag) return ;
	if(u==v)
	{
		printf(\"%d\\n\",ans);
		return ;
	}
	vis[u]=1;
	for(int i=0;i<pp[u].size();i++)
	{
		dfs(pp[u][i].x,ans+pp[u][i].w,v);
	}
	vis[u]=0;
}

int main()
{
	int t;
	scanf(\"%d\",&t);
	while(t--)
	{
		scanf(\"%d%d\",&n,&m);
		for(int i=0;i<n-1;i++)
		{
			int a,b,c;
			scanf(\"%d%d%d\",&a,&b,&c);
			p.x=b;
			p.w=c;
			pp[a].push_back(p);
			p.x=a;
			p.w=c;
			pp[b].push_back(p);
		}
		while(m--)
		{
			int a,b;
			scanf(\"%d%d\",&a,&b);
			memset(vis,0,sizeof(vis));
			flag=0;
			dfs(a,0,b);
		}
	}
	
	
}

努力加油a啊，(^o)/~