hud 2586 How far away ？

题目连接

http://acm.hdu.edu.cn/showproblem.php?pid=2586  

How far away ？

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

tarjan离线求LCA。。

#include <bits/stdc++.h>
using namespace std;
const int N = 40010;
struct Tarjan_Lca {
    bool vis[N];
    vector<pair<int, int>> A, Q[N];
    int tot, par[N], ans[210], head[N], dist[N];
    struct edge { int to, w, next; }G[N << 1];
    inline void init(int n) {
        A.clear();
        for(int i = 0; i < n + 2; i++) {
            par[i] = i;
            vis[i] = false;
            head[i] = -1;
            dist[i] = 0;
            Q[i].clear();
        }
    }
    inline void add_edge(int u, int v, int w) {
        G[tot] = { v, w, head[u] }, head[u] = tot++;
        G[tot] = { u, w, head[v] }, head[v] = tot++;
    }
    inline void built(int n, int m) {
        int u, v, w;
        while(n-- > 1) {
            scanf("%d %d %d", &u, &v, &w);
            add_edge(u, v, w);
        }
        for(int i = 0; i < m; i++) {
            scanf("%d %d", &u, &v);
            A.push_back(pair<int, int>(u, v));
            Q[u].push_back(pair<int, int>(v, i));
            Q[v].push_back(pair<int, int>(u, i));
        }
    }
    inline int find(int x) {
        while(x != par[x]) {
            x = par[x] = par[par[x]];
        }
        return x;
    }
    inline void tarjan(int u, int fa) {
        for(int i = head[u]; ~i; i = G[i].next) {
            edge &e = G[i];
            if(e.to == fa) continue;
            dist[e.to] = dist[u] + e.w;
            tarjan(e.to, u);
            vis[e.to] = true;
            par[e.to] = u;
        }
        for(auto &r: Q[u]) {
            if(vis[r.first]) ans[r.second] = find(r.first);
        }
    }
    inline void solve(int n, int m) {
        init(n);
        built(n, m);
        tarjan(1, 1);
        for(int i = 0; i < m; i++) {
            printf("%d\n", dist[A[i].first] + dist[A[i].second] - 2 * dist[ans[i]]);
        }
    }
}go;
int main() {
#ifdef LOCAL
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w+", stdout);
#endif
    int t, n, m;
    scanf("%d", &t);
    while(t--) {
        scanf("%d %d", &n, &m);
        go.solve(n, m);
    }
    return 0;
}

转载于:https://www.cnblogs.com/GadyPu/p/4992920.html