Hdu 2586(LCA)

题目链接

How far away ？

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5491    Accepted Submission(s): 2090

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

Accepted Code:

/*************************************************************************
    > File Name: 2586.cpp
    > Author: Stomach_ache
    > Mail: sudaweitong@gmail.com
    > Created Time: 2014年09月03日 星期三 09时53分59秒
    > Propose: 
 ************************************************************************/
#pragma comment(linker, "/STACK:1024000000, 1024000000")
#include <cmath>
#include <string>
#include <cstdio>
#include <vector>
#include <fstream>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
/*Let's fight!!!*/

typedef pair<int, int> pii;
const int MAX_V = 40050;
const int MAX_LOG_V = 20;
vector<pii> G[MAX_V];
int root;

int depth[MAX_V];
int fa[MAX_LOG_V][MAX_V];
int dis[MAX_V];

void dfs(int u, int p, int d) {
      fa[0][u] = p;
      depth[u] = d;
    for (int i = 0; i < G[u].size(); i++) {
          int v = G[u][i].first, w = G[u][i].second;
          if (v != p) {
            dis[v] = dis[u] + w;
              dfs(v, u, d + 1);
        }
    }
}

void init(int V) {
      dis[1] = 0;
    root = 1;
      dfs(root, -1, 0);

    for (int k = 0; k + 1 < MAX_LOG_V; k++) {
          for (int v = 1; v <= V; v++) {
              if (fa[k][v] < 0) fa[k + 1][v] = -1;
            else fa[k + 1][v] = fa[k][fa[k][v]];
        }
    }
}

int lca(int u, int v) {
      if (depth[u] > depth[v]) swap(u, v);
    for (int k = 0; k < MAX_LOG_V; k++) {
          if ((depth[v] - depth[u]) >> k & 1) 
              v = fa[k][v];
    }

    if (u == v) return u;
    for (int k = MAX_LOG_V - 1; k >= 0; k--) {
          if (fa[k][u] != fa[k][v]) {
              u = fa[k][u];
            v = fa[k][v];
        }
    }
    return fa[0][u];
}

int main(void) {
      ios_base::sync_with_stdio(false);
      int test;
    cin >> test;
    while (test--) {
          int n, m;
        cin >> n >> m;
        for (int i = 1; i <= n; i++) G[i].clear();

        for (int i = 1; i < n; i++) {
              int u, v, w;
            cin >> u >> v >> w;
            G[u].push_back(pii(v, w));
            G[v].push_back(pii(u, w));
        }

        init(n);
        while (m--) {
              int u, v;
            cin >> u >> v;
            cout << dis[u] + dis[v] - dis[lca(u, v)] * 2 << endl;
        }
    }

    return 0;
}