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就可以過，vector真是個好東西。總感覺這個題目出水了，40000的空間肯定不能開個二維空間，vector真的可以大大的減少空間。
代碼如下：

#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里面存結(jié)構(gòu)體 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)/~

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