HDU 2586 How far away ？

How far away ？

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 17350    Accepted Submission(s): 6696

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

 

 

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LCA的大水题。

#include<iostream>
#include<map>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define MAXN 500000+15
using namespace std;
int T,n,m,tot;
int dad[MAXN*4],deep[MAXN*4],size[MAXN*4],top[MAXN*4],length[MAXN*4];
int to[MAXN*5],from[MAXN*5],net[MAXN*5],cap[MAXN*5];
void add(int u,int v,int w){
    to[++tot]=v;net[tot]=from[u];from[u]=tot;cap[tot]=w;
    to[++tot]=u;net[tot]=from[v];from[v]=tot;cap[tot]=w;
}
void dfs(int x){
    size[x]=1;
    deep[x]=deep[dad[x]]+1;
    for(int i=from[x];i;i=net[i])
        if(dad[x]!=to[i]){
            dad[to[i]]=x;
            length[to[i]]=length[x]+cap[i];
            dfs(to[i]);
            size[x]+=size[to[i]];
        }
}
void dfs1(int x){
    int t=0;
    if(!top[x])    top[x]=x;
    for(int i=from[x];i;i=net[i])
        if(dad[x]!=to[i]&&size[t]<size[to[i]])
            t=to[i];
    if(t){
        top[t]=top[x];
        dfs1(t);
    }
    for(int i=from[x];i;i=net[i])
        if(dad[x]!=to[i]&&t!=to[i])
            dfs1(to[i]);
}
int lca(int x,int y){
    for(;top[x]!=top[y];){
        if(deep[top[x]]<deep[top[y]])
            swap(x,y);
        x=dad[top[x]];
    }
    if(deep[x]>deep[y])
        swap(x,y);
    return x;
}
int main(){
    cin>>T;
    while(T--){
        tot=0; 
        memset(to,0,sizeof(to));
        memset(cap,0,sizeof(cap));
        memset(net,0,sizeof(net));
        memset(top,0,sizeof(top));
        memset(dad,0,sizeof(dad));
        memset(from,0,sizeof(from));
        memset(size,0,sizeof(size));
        memset(deep,0,sizeof(deep));
        memset(length,0,sizeof(length));
        scanf("%d%d",&n,&m);
        for(int i=1;i<n;i++){
            int u,v,z;
            scanf("%d%d%d",&u,&v,&z);
            add(u,v,z);
        }
        dfs(1);
        dfs1(1);
        for(int i=1;i<=m;i++){
            int u,v;
            scanf("%d%d",&u,&v);
            int l=lca(u,v);
            cout<<length[u]+length[v]-2*length[l]<<endl;
        }
    }
}