hdu2586How far away ？

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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

#include<map>
#include<string>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#include<bitset>
#include<climits>
#include<list>
#include<iomanip>
#include<stack>
#include<set>
using namespace std;
int head[40000+10],tail;
struct Edge
{
    int val,to,next;
}edge[2*40000+10];
void add(int from,int to,int val)
{
    edge[tail].val=val;
    edge[tail].to=to;
    edge[tail].next=head[from];
    head[from]=tail++;
}
int step,fst[40000+10],dth[40000*2+10],dis[40000+10],point[40000*2+10];
bool vis[40000+10];
void dfs(int from,int dep)
{
    vis[from]=1;
    fst[from]=++step;
    dth[step]=dep;
    point[step]=from;
    for(int i=head[from];i!=-1;i=edge[i].next)
    {
        int to=edge[i].to,val=edge[i].val;
        if(!vis[to])
        {
            dis[to]=dis[from]+val;
            dfs(to,dep+1);
            point[++step]=from;
            dth[step]=dep;
        }
    }
}
int dp[40000*2+10][20];
void create(int n)
{
    n*=2;
    for(int i=0;i<n;i++)
        dp[i][0]=i;
    for(int j=1;(1<<j)<n;j++)
        for(int i=0;i+(1<<j)-1<n;i++)
        {
            int one=dp[i][j-1],two=dp[i+(1<<j-1)][j-1];
            dp[i][j]=dth[one]<dth[two]?one:two;
        }
}
int seek(int l,int r)
{
    int k,len;
    for(k=0,len=r-l+1;(1<<k+1)<=len;k++);
	int one=dp[l][k],two=dp[r-(1<<k)+1][k];
    return dth[one]<dth[two]?point[one]:point[two];
}
int cnt(int l,int r)
{
    int one=fst[l],two=fst[r];
    if(one>two)
        swap(one,two);
    return dis[l]+dis[r]-2*dis[seek(one,two)];
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n,m;
        scanf("%d%d",&n,&m);
        tail=0;
        memset(head,-1,sizeof(head));
        for(int i=1;i<n;i++)
        {
            int from,to,val;
            scanf("%d%d%d",&from,&to,&val);
            add(from,to,val);
            add(to,from,val);
        }
        dis[1]=0;
        memset(vis,0,sizeof(vis));
        step=-1;
        dfs(1,0);
        create(n);
        while(m--)
        {
            int l,r;
            scanf("%d%d",&l,&r);
            printf("%d\n",cnt(l,r));
        }
    }
}