hdu 2586 How far away ? ( 离线 LCA ， tarjan )

How far away ？

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10312    Accepted Submission(s): 3743

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

 

#include <bits/stdc++.h>
using namespace std ;
const int N = 100010 ;

typedef pair<int,int> pii ;
#define X first
#define Y second
int n , m ;
int eh[N] , nxt[N<<1] , et[N<<1] , ew[N<<1] , tot ;
int fa[N] , vis[N] , anc[N] ;
int ux[N] , vx[N] , wx[N] ;

 vector<pii>qry[N] ;

void init() {

    for( int i = 0 ; i <= n ; ++i ) {
        qry[i].clear() ;
        vis[i] = 0 ;
    }

    tot = 0 ;
    memset( eh , -1 , sizeof eh ) ;

}

void addedge( int u , int v , int w ) {
    et[tot] = v , nxt[tot] = eh[u] , ew[tot] = w , eh[u] = tot++ ;
}

int find(int x){ return fa[x] = ( fa[x]==x?x:find(fa[x]));}
void un( int x , int y ) {
    x = find(x);
    y = find(y);
    if(x==y)return ;
    fa[y]=x;
}
void dfs1( int u , int pre ) {

    vis[u] = 1 ;
    fa[u] = u ;
    for( int i = eh[u] ; ~i ; i = nxt[i] ) {
        int v = et[i] ; if( v == pre ) continue ;
        dfs1( v , u ) ;
        un( u , v );
    }

    int siz = qry[u].size() ;
    for( int i = 0 ; i < siz ; ++i ) {
        if( vis[qry[u][i].X] ) {
            anc[qry[u][i].Y] = find( qry[u][i].X );
        }
    }
}

int dis[N] ;

void dfs2( int u , int pre , int d ) {
    dis[u] = d ;
    for( int i = eh[u] ; ~i ; i = nxt[i] ) {
        int v = et[i] ; if( v == pre ) continue ;
        dfs2( v , u , d + ew[i] ) ;
    }
}

int main() {
    #ifdef LOCAL
        freopen("in.txt","r",stdin);
    #endif // LOCAL
    int _ ; scanf("%d",&_) ;
    while( _-- ) {

        scanf("%d%d",&n,&m) ;
        init() ;
        for( int i = 1 ; i < n ; ++i ) {
            int u , v , w ;
            scanf("%d%d%d",&u,&v,&w) ;
            addedge( u , v , w ) ;
            addedge( v , u , w ) ;
        }

        for( int i = 0 ; i < m ; ++i ) {
            scanf("%d%d",&ux[i],&vx[i]) ;
            qry[ ux[i] ].push_back( pii( vx[i] , i ) ) ;
            qry[ vx[i] ].push_back( pii( ux[i] , i ) ) ;
        }

        dfs1( 1 , 0 ) ;
        dfs2( 1 , 0 , 0 ) ;

        for( int i = 0 ; i < m ; ++i ) {
            printf("%d\n",dis[ux[i]]+dis[vx[i]]-2*dis[anc[i]]);
        }
    }
    return 0 ;
}