hdu 2874 森林中处理最近公共祖先问题

来源：互联网发布：期货程序化交易软件编辑：程序博客网时间：2024/05/01 21:14

注意两个问题,第一个是在森林中,所以需要判断两个点是否是在一棵树中,第二个是求取路径时可以借助到根的路径长度来求取,类似于前缀和的一种优化

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <vector>#define MAX 20007using namespace std;int n , m  , c , u , v , w ;struct{    int v , next , w;}e[MAX<<1];struct {    int v , next, index; }q[2000007];int cc = 0;int head1[MAX];int head2[MAX];int fa[MAX];int find ( int x ){    return fa[x] == x ? x : fa[x] = find ( fa[x] );}void add1 ( int u , int v , int w ){    e[cc].v = v;    e[cc].w = w;    e[cc].next = head1[u];    head1[u] = cc++;}void add2 ( int u , int v , int i ){    q[cc].v = v;    q[cc].next = head2[u];    q[cc].index = i;    head2[u] = cc++;}int vis[MAX];int root[MAX];int dis[MAX];int yes[1000007];void init (){    memset ( head1 , -1 , sizeof ( head1 ) );    memset ( head2 , -1 , sizeof ( head2 ) );    cc = 0;    for ( int i = 1 ; i <= n+1 ; i++ ) fa[i] = i;    memset ( vis , 0 , sizeof ( vis ) );    memset ( dis , 0 , sizeof ( dis ) );    memset ( root , 0 , sizeof ( root ) );} void lca ( int u , int p , int r ){    vis[u] = 1;    root[u] = r;    for ( int i = head1[u] ; i != -1 ; i = e[i].next )    {        int v = e[i].v;        if ( v == p ) continue;        dis[v] = dis[u] + e[i].w;        lca ( v , u , r );        fa[v] = u;    }    for ( int i = head2[u] ; i != -1 ; i = q[i].next )    {        int v = q[i].v;        if ( root[v] != r ) continue;        int l = find ( v );        yes[q[i].index] = dis[u] + dis[v] - 2*dis[l];    }}int main ( ){    while ( ~scanf ( "%d%d%d" , &n , &m , &c ) )    {        init ( );        for ( int i = 0 ; i < m ; i++ )        {            scanf ( "%d%d%d" , &u ,&v , &w );            add1 ( u , v , w );            add1 ( v , u , w );        }        cc = 0;        for ( int i = 0 ; i < c ; i++ )        {            scanf ( "%d%d" , &u , &v );            add2 ( u , v , i );            add2 ( v , u , i );        }        memset ( yes , -1 , sizeof ( yes ) );        for ( int i = 1 ; i <= n ; i++ )            if ( !vis[i] ) lca ( i , -1 , i );        for ( int i = 0 ; i < c ; i ++ )            if ( yes[i] == -1 )                printf ( "Not connected\n" );             else printf ( "%d\n" , yes[i] );    }    return 0;}

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