hdu 2586 How far away ？ Lca的模板了、、

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 <stdio.h>
#include <string.h>
#include <algorithm>
#include <vector>
#include <stack>
#include <iostream>
#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<n;i++)
       {
           scanf("%d %d %d",&a,&b,&k);
           edge[e1].to=b;
           edge[e1].dis=k;
           edge[e1].next=head[a];
           head[a]=e1;
           e1++;

           edge[e1].to=a;
           edge[e1].dis=k;
           edge[e1].next=head[b];
           head[b]=e1;
           e1++;
       }
       for(i=1;i<=m;i++)
       {
           scanf("%d %d",&a,&b);
           qedge[e2].to=b;
           qedge[e2].next=qhead[a];
           qhead[a]=e2;
           e2++;

           qedge[e2].to=a;
           qedge[e2].next=qhead[b];
           qhead[b]=e2;
           e2++;
       }
       for(i=1;i<=n;i++)
        f[i]=i,visit[i]=false;
       dis[1]=0;
       LCA(1);
       for(i=0;i<e2;i+=2)
        {
            printf("%I64d\n",qedge[i].dis);
        }
    }
    return 0;
}