hdu 2586 How far away ? （LCA 离线tarjan）

How far away ？

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 18126    Accepted Submission(s): 7037

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

23 21 2 103 1 151 22 32 21 2 1001 22 1

 

Sample Output

1025100100

 

Source

ECJTU 2009 Spring Contest 

复习一下图论的知识。。

关于LCA的离线算法可以看一下这个。。。 http://www.cnblogs.com/JVxie/p/4854719.html。 写的挺好的（我懒得写

看了好多的博客。。的模板都不是很喜欢其风格。。

#include <iostream>#include <string.h>#include <algorithm>#include <stdio.h>#include <utility>#include <vector>using namespace std;const int maxn = 80305;int fa[maxn], dist[maxn], visited[maxn];int head[maxn];//树的边集struct Node {    int u, v, w, next;}node[maxn];int num;struct Edge {    int u, v, lca;};//查找的点与点的集合vector<Edge> edges;int find(int x) {    if (x == fa[x])        return x;    return fa[x] = find(fa[x]);}void dfs(int u) {    fa[u] = u;    visited[u] = 1;    for (int i = 0; i < edges.size(); ++i) {        //边中的某点与当前访问的点有关，且另一个点已经被访问过了        if (edges[i].u == u && visited[edges[i].v]) {            //找当前的公共祖先            edges[i].lca = find(edges[i].v);        } else if (edges[i].v == u && visited[edges[i].u]) {            edges[i].lca = find(edges[i].u);        }    }        for (int i = head[u]; i != -1; i = node[i].next) {        int v = node[i].v;        int w = node[i].w;        if (!visited[v]) {            dist[v] = dist[u] + w;            dfs(v);            //回溯时标记父节点，才能在当前dfs中找到公共祖先            fa[v] = u;        }    }}void init() {    num = 0;    memset(head, -1, sizeof(head));    memset(dist, 0, sizeof(dist));    edges.clear();    memset(visited, 0, sizeof(visited));}void add(int u, int v, int w) {    node[num] = {u, v, w, head[u]};    head[u] = num++;    node[num] = {v, u, w, head[v]};    head[v] = num++;}int main() {        //freopen("in.txt", "r", stdin);    ios::sync_with_stdio(false);    std::cin.tie(0);    //给vector提前扩大空间，减少开销    edges.reserve(250);    int t;    cin >> t;    while (t--) {        init();        int n, m;        cin >> n >> m;        for (int i = 1; i < n; ++i) {            int u, v, w;            cin >> u >> v >> w;            add(u, v, w);        }                for (int i = 0; i < m; ++i) {            int u, v;            cin >> u >> v;            edges.push_back({u, v, 0});        }                dfs(1);                for (int i = 0; i < m; ++i) {            cout << dist[edges[i].u] + dist[edges[i].v] - 2 * dist[edges[i].lca] << endl;        }    }}