链接:

How far away ？

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11204    Accepted Submission(s): 4079

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

一个村子里有n个房子，这n个房子用n-1条路连接起来，接下了有m次询问，每次询问两个房子a,b之间的距离是多少。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <algorithm>
#include <set>
using namespace std;
typedef long long ll;
typedef unsigned long long Ull;
#define MM(a,b) memset(a,b,sizeof(a));
const double eps = 1e-10;
const int  inf =0x7f7f7f7f;
const double pi=acos(-1);
const int maxn=40000;

struct Edge{
    int to,w;
    Edge(int a,int b):to(a),w(b){};
    Edge(){};
}e[maxn+10];

struct node{
   int to,id;
};

vector<Edge> G[maxn+10];
vector<node> mp[maxn+10];
int vis[maxn+10],query[maxn+10][4],par[maxn+10],dis[maxn+10];

int findr(int u)
{
    if(par[u]!=u)
        par[u]=findr(par[u]);
    return par[u];
}

void unite(int u,int v)
{
    int ru=findr(u);
    int rv=findr(v);
    if(ru!=rv) par[rv]=ru;
}

void tarjan(int u,int val)
{
    vis[u]=1;
    dis[u]=val;

    for(int i=0;i<mp[u].size();i++)
    {
        int v=mp[u][i].to;
        int id=mp[u][i].id;
        if(vis[v])
            query[id][2]=findr(v);
    }

    for(int i=0;i<G[u].size();i++)
    {
        int v=G[u][i].to;
        if(vis[v]) continue;
        tarjan(v,val+G[u][i].w);
        unite(u,v);
    }

}

int main()
{
    int cas,n,op,u,v,w;
    scanf("%d",&cas);
    while(cas--)
    {
        scanf("%d %d",&n,&op);

        for(int i=1;i<=n;i++)
        {
            G[i].clear();
            mp[i].clear();
            vis[i]=0;
            par[i]=i;
        }

        for(int i=1;i<=n-1;i++)
        {
            scanf("%d %d %d",&u,&v,&w);
            G[u].push_back(Edge(v,w));
            G[v].push_back(Edge(u,w));
        }

        for(int i=1;i<=op;i++)
        {
            scanf("%d %d",&u,&v);
            query[i][0]=u;
            query[i][1]=v;
            mp[u].push_back((node){v,i});
            mp[v].push_back((node){u,i});
        }

        tarjan(1,0);

        for(int i=1;i<=op;i++)
        {
            int u=query[i][0];
            int v=query[i][1];
            int r=query[i][2];
            printf("%d\n",dis[u]+dis[v]-2*dis[r]);
        }
    }
    return 0;
}

分析：用的lca

1.因为有n-1条边，任意两点之间可达，那么就是一棵树；

2，假设当前是询问u和v两个点，他们的LCA是r,dis[]数组代表从根到点的距离，那么u和v两点之间的距离就是dis[u]+dis[v]-2*dis[r]；所以只需要用Tarjan求一求询问的点对的LCA并且记录一下每个点到根（随便哪个点做根都可以）的距离就可以了