hdu2586（How far away ？）

                                                   How far away ？

                                     Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
                                             Total Submission(s): 2561    Accepted Submission(s): 946

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 经典问题：求最近公共祖先。

//Accepted	2586	125MS	4056K	1713 B	C++

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <string>

#include <map>

#include <vector>

using namespace std;

const int maxn = 40100;

int f[maxn];

int d[maxn];            //保存每个节点的深度。



vector<int> a[maxn];

map<int, int> h;        //保存每个节点到其父亲边的距离。

int n, m;



void getDep(int num, int dep) {

    d[num] = dep;

    vector<int>::iterator it;

    it = a[num].begin();

    for(; it < a[num].end(); ++it) {

        getDep(*it, dep+1);

    }

}



int work(int a, int b) {

    int s1 = 0;

    int s2 = 0;

    while(a!=b) {

        while(d[a]<d[b]){

            s1 += h[b];

            b = f[b];

        }

        while(d[a]>d[b]) {

            s2 += h[a];

            a = f[a];

        }

        if(d[a]==d[b]&&a!=b) {

            s1 += h[b];

            b = f[b];

        }

    }

    return s1+s2;

}



void init() {

    for(int i = 0; i < n; i++) {

        a[i].clear();

    }

    h.clear();

}



int main()

{

    int T;

    int u;

    scanf("%d", &T);

    int from, to, w;

    for(u = 0; u < T; u++) {

        scanf("%d%d", &n, &m);

        for(int i = 0; i < n-1; i++) {

            scanf("%d%d%d", &from, &to, &w);

            a[from].push_back(to);

            f[to] = from;

            h[to] = w;//保存每个节点到其父亲的距离

        }

        getDep(1, 0);

        //query

        for(int i = 0; i < m; i++) {

            scanf("%d%d", &from, &to);

            int res = work(from, to);

            cout << res << endl;

        }

        init();

    }

    return 0;

}