【例题题解】Cow Contest：floyed传递闭包

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题目描述

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 ≤ A ≤ N; 1 ≤ B ≤ N; A ≠ B), 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.

FJ的N(1 <= N <= 100)头奶牛们最近参加了场程序设计竞赛:)。在赛场上，奶牛们按1…N依次编号。每头奶牛的编程能力不尽相同，并且没有哪两头奶牛的水平不相上下，也就是说，奶牛们的编程能力有明确的排名。整个比赛被分成了若干轮，每一轮是两头指定编号的奶牛的对决。如果编号为A的奶牛的编程能力强于编号为B的奶牛(1 <= A <= N; 1 <= B <= N; A != B) ，那么她们的对决中，编号为A的奶牛总是能胜出。 FJ想知道奶牛们编程能力的具体排名，于是他找来了奶牛们所有 M(1 <= M <= 4,500)轮比赛的结果，希望你能根据这些信息，推断出尽可能多的奶牛的编程能力排名。比赛结果保证不会自相矛盾。

Solution

这道题用 $f l o y e d$ &传递闭包的思路： $f [i] [j]$ 表示i是否能打败j。
三重循环枚举， $f[i][j]|=f[i][k]\&\&f[k][j]$
表示第i头牛能打败的牛所打败的牛一定被第i头牛打败。
即：i打败j，j打败k，则i打败k。
然后对于每一只奶酒，只要打败它或它能打败的奶牛个数为n-1说明排名能够确定。
代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
const int N=300;
int n,m,ans=0;
bool f[N][N];
inline void read(int &s)
{char c=getchar();s=0;while (c<'0' || c>'9') c=getchar();while (c>='0' && c<='9') s=s*10+c-48,c=getchar();
}
int main(void)
{read(n),read(m);for (int i=1;i<=m;++i){int a,b;read(a),read(b);f[a][b]=1;}for (int k=1;k<=n;++k)for (int i=1;i<=n;++i)for (int j=1;j<=n;++j)f[i][j]|=f[i][k]&&f[k][j];for (int i=1;i<=n;++i){int sum=0;for (int j=1;j<=n;++j)if (j!=i && (f[j][i] || f[i][j])) sum++;if (sum==n-1) ans++;}cout<<ans<<endl;return 0;
}