LCA距离

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

 

 

n个定点，每个点之间有距离，q次询问，求两点之间的最短路径，和最小生成树有所不同的是n个定点只有n-1个边使之相通，没有连成圆。

用LCA做时，连点之间的距离等于根节点到两点距离之和 减去根节点到点v点的祖先的距离的两倍 即ans[query[u][i].w] = dis[u]+dis[v]-2*dis[findset(v)];

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <vector>
using namespace std;
const int N = 40010;
struct Nod{
    int to, w;
    Nod(int a = 0, int b = 0){
        to = a; w = b;
    }
};
int ans[N];
int par[N],vis[N],mark[N],dis[N];
int n,m;
vector<Nod> mp[N];
vector<Nod> query[N];
void init(){
    for(int i = 0; i <= n; i ++){
        mp[i].clear();
        query[i].clear();
        vis[i] = mark[i] = 0;
        par[i] = i;
    }
    memset(ans,-1,sizeof(ans));
}
int findset(int x){
    if(x != par[x]) par[x] = findset(par[x]);
    return par[x];
}

void dfs(int u){
    for(int i = 0; i < query[u].size(); i ++){
        int v = query[u][i].to;
        if(vis[v]&&ans[query[u][i].w]==-1&&!mark[findset(v)]){
            ans[query[u][i].w] = dis[u]+dis[v]-2*dis[findset(v)];
        }
    }
    for(int i = 0; i < mp[u].size(); i++){
        int v = mp[u][i].to;
        if(!vis[v]){
            vis[v] = 1;
            dis[v] = dis[u]+mp[u][i].w;
            dfs(v);
            par[v] = u;
        }
    }
}

int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&m);
        init();
        int x, y, z;
        for(int i = 0; i < n-1; i ++){
            scanf("%d%d%d",&x,&y,&z);
            mp[x].push_back(Nod(y,z));
            mp[y].push_back(Nod(x,z));
        }
        for(int i = 1; i <= m; i ++){
            scanf("%d%d",&x,&y);
            query[x].push_back(Nod(y,i));
            query[y].push_back(Nod(x,i));
        }
        for(int i = 1; i <= n; i ++){            
            if(!vis[i]){
                printf("%d+++
",i);
                vis[i] = 1;
                dis[i] = 0;
                dfs(i);
                mark[i] = 1;
            }
        }
        for(int i = 1; i <= m; i ++){
            printf("%d
",ans[i]);
        }
    }
    return 0;
}