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;
}