• 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
    ",dis[ux[i]]+dis[vx[i]]-2*dis[anc[i]]);
            }
        }
        return 0 ;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/hlmark/p/5137549.html
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