前往小程序,Get更优阅读体验!
立即前往
首页
学习
活动
专区
工具
TVP
发布
社区首页 >专栏 >poj-3660-cows contest(不懂待定)

poj-3660-cows contest(不懂待定)

作者头像
瑾诺学长
发布2018-09-21 16:10:31
3650
发布2018-09-21 16:10:31
举报
文章被收录于专栏:专注研发

Description

N (1 ≤ N ≤ 100) cows, conveniently numbered 1..N, are participating in a programming contest. As we all know, some cows code better than others. Each cow has a certain constant skill rating that is unique among the competitors.

The contest is conducted in several head-to-head rounds, each between two cows. If cow A has a greater skill level than cow B (1 ≤ AN; 1 ≤ BN; AB), then cow A will always beat cow B.

Farmer John is trying to rank the cows by skill level. Given a list the results of M (1 ≤ M ≤ 4,500) two-cow rounds, determine the number of cows whose ranks can be precisely determined from the results. It is guaranteed that the results of the rounds will not be contradictory.

Input

* Line 1: Two space-separated integers: N and M * Lines 2..M+1: Each line contains two space-separated integers that describe the competitors and results (the first integer, A, is the winner) of a single round of competition: A and B

Output

* Line 1: A single integer representing the number of cows whose ranks can be determined

Sample Input

代码语言:javascript
复制
5 5
4 3
4 2
3 2
1 2
2 5
分析:有人说叫闭包传递。这题就是用一种类似于floyd的算法,
开始时,如果a胜b则由a到b连一条边。这样所有a能走到的点都是排名在a以后的。
所有能走到a的点都是排名在a以前的。用floyd,求出每个点能到达的点。
如果有一个点,排名在它之前的和排名在它之后的点之和为n-1,那么它的排名就是确定的。
代码语言:javascript
复制
#include<iostream>
using namespace std ;
int main()
{
    int N,M,a,b;
    cin>>N>>M;
    int aa[110][110]={0};
      while(M--)
    {
        cin>>a>>b;
        aa[a][b]=1;
    }

    for(int i=1;i<=N;i++)
        for(int j=1;j<=N;j++)
        for(int k=1;k<=N;k++)
    {
        if(aa[j][i]&&aa[i][k])
            aa[j][k]=1;
    }
    int ans=0;
    for(int i=1;i<=N;i++)
    {
        int tmp=0;
        for(int j=1;j<=N;j++)
        tmp+=aa[i][j]+aa[j][i];
        if(tmp==N-1)ans++;


    }

  cout<<ans<<endl;
    return 0;
}
本文参与 腾讯云自媒体同步曝光计划,分享自作者个人站点/博客。
原始发表:2016-03-22 ,如有侵权请联系 cloudcommunity@tencent.com 删除

本文分享自 作者个人站点/博客 前往查看

如有侵权,请联系 cloudcommunity@tencent.com 删除。

本文参与 腾讯云自媒体同步曝光计划  ,欢迎热爱写作的你一起参与!

评论
登录后参与评论
0 条评论
热度
最新
推荐阅读
领券
问题归档专栏文章快讯文章归档关键词归档开发者手册归档开发者手册 Section 归档