zoukankan      html  css  js  c++  java
  • BZOJ 1648: [Usaco2006 Dec]Cow Picnic 奶牛野餐( dfs )

    直接从每个奶牛所在的farm dfs , 然后算一下..

    ----------------------------------------------------------------------------------------

    #include<cstdio>
    #include<algorithm>
    #include<cstring>
    #include<iostream>
    #include<vector>
     
    #define rep( i , n ) for( int i = 0 ; i < n ; ++i )
    #define clr( x , c ) memset( x , c , sizeof( x ) )
     
    using namespace std;
     
    const int maxn = 1000 + 5;
    const int maxk = 100 + 5;
     
    vector< int > G[ maxn ];
     
    int a[ maxk ];
    bool vis[ maxn ];
    int cnt[ maxn ];
     
    void dfs( int x ) {
    vis[ x ] = true;
    cnt[ x ]++;
    rep( i , G[ x ].size() ) {
    int to = G[ x ][ i ];
    if( vis[ to ] ) continue;
    dfs( to );
    }
    }
     
    int main() {
    freopen( "test.in" , "r" , stdin );
    int k , n , m;
    cin >> k >> n >> m;
    rep( i , n ) G[ i ].clear();
    rep( i , k ) 
       scanf( "%d" , a + i ) , a[ i ]--;
    while( m-- ) {
    int u , v;
    scanf( "%d%d" , &u , &v );
    u-- , v--;
    G[ u ].push_back( v );
    }
    clr( cnt , 0 );
    rep( i , k ) 
       clr( vis , 0 ) , dfs( *( a + i ) );
       
    int ans = 0;
    rep( i , n ) if( cnt[ i ] == k )
       ans++;
    cout << ans << " ";
    return 0;
    }

    ----------------------------------------------------------------------------------------

    1648: [Usaco2006 Dec]Cow Picnic 奶牛野餐

    Time Limit: 5 Sec  Memory Limit: 64 MB
    Submit: 456  Solved: 283
    [Submit][Status][Discuss]

    Description

    The cows are having a picnic! Each of Farmer John's K (1 <= K <= 100) cows is grazing in one of N (1 <= N <= 1,000) pastures, conveniently numbered 1...N. The pastures are connected by M (1 <= M <= 10,000) one-way paths (no path connects a pasture to itself). The cows want to gather in the same pasture for their picnic, but (because of the one-way paths) some cows may only be able to get to some pastures. Help the cows out by figuring out how many pastures are reachable by all cows, and hence are possible picnic locations.

      K(1≤K≤100)只奶牛分散在N(1≤N≤1000)个牧场.现在她们要集中起来进餐.牧场之间有M(1≤M≤10000)条有向路连接,而且不存在起点和终点相同的有向路.她们进餐的地点必须是所有奶牛都可到达的地方.那么,有多少这样的牧场呢?

    Input

    * Line 1: Three space-separated integers, respectively: K, N, and M * Lines 2..K+1: Line i+1 contains a single integer (1..N) which is the number of the pasture in which cow i is grazing. * Lines K+2..M+K+1: Each line contains two space-separated integers, respectively A and B (both 1..N and A != B), representing a one-way path from pasture A to pasture B.

     1行输入KNM.接下来K行,每行一个整数表示一只奶牛所在的牧场编号.接下来M行,每行两个整数,表示一条有向路的起点和终点

    Output

    * Line 1: The single integer that is the number of pastures that are reachable by all cows via the one-way paths.

        所有奶牛都可到达的牧场个数

    Sample Input

    2 4 4
    2
    3
    1 2
    1 4
    2 3
    3 4


    INPUT DETAILS:

    4<--3
    ^ ^
    | |
    | |
    1-->2

    The pastures are laid out as shown above, with cows in pastures 2 and 3.

    Sample Output

    2

    牧场3,4是这样的牧场.

    HINT

    Source

  • 相关阅读:
    iOS--通讯录、蓝牙、内购、GameCenter、iCloud、Passbook等系统服务开发汇总
    iOS-网络爬虫
    iOS-性能优化
    iOS开发——网络实用技术OC篇&网络爬虫-使用青花瓷抓取网络数据
    深入解析Linux内核及其相关架构的依赖关系
    详解Linux系统中的文件名和文件种类以及文件权限
    Linux系统中使用netcat命令的奇技淫巧
    Linux系统下强大的lsof命令使用宝典
    Linux下多线程下载工具MWget和Axel使用介绍
    Linux下针对路由功能配置iptables的方法详解
  • 原文地址:https://www.cnblogs.com/JSZX11556/p/4567211.html
Copyright © 2011-2022 走看看