How far away ？

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2522    Accepted Submission(s): 931

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

// 差不多就是Lca的模板了、、 dis[i]代表根节点到i的距离 #include 
     
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           #include 
          
            #define Max 40001 using namespace std; struct Node {     int to;     int next;     __int64 dis;//本来是 LCA }; int qhead[Max]; Node qedge[522]; int head[Max]; Node edge[Max<<1]; __int64 dis[Max]; int f[Max]; bool visit[Max]; int Find(int x) {     if(f[x]!=x) f[x]=Find(f[x]);     return f[x]; } void LCA(int u) {     visit[u]=true;     int e,v;     for(e=head[u];e!=-1;e=edge[e].next)     {         v=edge[e].to;         if(!visit[v])         {             dis[v]=dis[u]+edge[e].dis;             LCA(v);             f[v]=u;         }     }     for(e=qhead[u];e!=-1;e=qedge[e].next)     {         if(visit[qedge[e].to])         {             int lca=Find(qedge[e].to); //最近公共祖先             qedge[e].dis=dis[u]+dis[qedge[e].to]-(dis[lca]+dis[lca]);             qedge[e^1].dis= qedge[e].dis;         }     } } int main() {     int n,m;     int T;     scanf("%d",&T);     while(T--)     {        int i;        int a,b,k;        int e1=0,e2=0;        scanf("%d %d",&n,&m);        for(i=1;i<=n;i++)         qhead[i]=head[i]=-1;        for(i=1;i