HDU 2586 How far away ？ 离线lca模板题

How far away ？

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8712    Accepted Submission(s): 3047

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

 

Source

ECJTU 2009 Spring Contest

题意

给你一棵树，每次询问两点间的距离。

题解

跑一发离线lca，每次询问u,v，设c是u,v的lca，那么答案就是dis[u]+dis[v]-2*dis[c]，其中dis表示从根到节点的距离。详见代码：

#pragma comment(linker, "/STACK:102400000,102400000")
#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
#define MAX_N 40003
using namespace std;

bool vis[MAX_N];

struct edge {
public:
    int to;
    long long cost;

    edge(int t, long long c) : to(t), cost(c) { }

    edge() { }
};

vector<edge> G[MAX_N];
int q,n,m;

struct node {
public:
    int p, v;

    node(int pp, int vv) : p(pp), v(vv) { }

    node() { }
};

vector<node> Q[MAX_N];
int lca[MAX_N], ancestor[MAX_N];

int T, cas = ;

int father[MAX_N];
void init() {
    for (int i = ; i <= n; i++)
        father[i] = i;
}

int Find(int u) {
    if (u == father[u])return u;
    else return father[u] = Find(father[u]);
}

void unionSet(int u,int v) {
    int x = Find(u), y = Find(v);
    if (x == y)return;
    father[x] = y;
}

bool Same(int u,int v) {
    return Find(u) == Find(v);
}

long long dis[MAX_N];

void Tarjan(int u,int p) {
    for (int i = ; i < G[u].size(); i++) {
        int v = G[u][i].to;
        if (v == p)continue;
        dis[v] = dis[u] + G[u][i].cost;
        Tarjan(v, u);
        unionSet(u, v);
        ancestor[Find(u)] = u;
    }
    vis[u] = ;
    for (int i = ; i < Q[u].size(); i++) {
        int v = Q[u][i].v;
        if (vis[v])
            lca[Q[u][i].p] = ancestor[Find(v)];
    }
}

pair<int, int> qu[MAX_N];

int main() {
    //cin.sync_with_stdio(false);
    scanf("%d", &T);
    while (T--) {
        scanf("%d%d", &n, &q);
        m = n - ;
        init();
        memset(vis, , sizeof(vis));
        memset(ancestor, , sizeof(ancestor));
        memset(dis, , sizeof(dis));
        memset(lca, , sizeof(lca));
        for (int i = ; i <= n; i++)G[i].clear();
        for (int i = ; i <= n; i++)Q[i].clear();

        for (int i = ; i < m; i++) {
            int u, v;
            long long c;
            scanf("%d%d%I64d", &u, &v, &c);
            G[u].push_back(edge(v, c));
            G[v].push_back(edge(u, c));
        }

        for (int i = ; i < q; i++) {
            int u, v;
            scanf("%d%d", &u, &v);
            qu[i] = make_pair(u, v);
            Q[u].push_back(node(i, v));
            Q[v].push_back(node(i, u));
        }
        Tarjan(, );

        for (int i = ; i < q; i++)
            printf("%I64d\n", dis[qu[i].first] + dis[qu[i].second] -  * dis[lca[i]]);
    }
    return ;
}