nyoj 211 Cow Contest 【最短路&&floyd】



Cow Contest

时间限制：1000 ms | 内存限制：65535 KB

难度：4

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.

输入* 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

There are multi test cases.The input is terminated by two zeros.The number of test cases is no more than 20.输出For every case:

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

5 5
4 3
4 2
3 2
1 2
2 5
0 0

样例输出

2

分析：

简单的floyd算法。

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#define mem(x,y) memset(x,y,sizeof(x))
using namespace std;
int cow[110][110];

void floyd(const int &n)
{
for(int k=1;k<=n;k++)
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
{
cow[i][j]=cow[i][j]||(cow[i][k]&&cow[k][j]);
}
}

int main()
{
int n,m;
while(scanf("%d%d",&n,&m)&&n&&m)
{
int u,v;
memset(cow,0,sizeof(cow));
for(int i=0;i<m;i++)
{
scanf("%d%d",&u,&v);
cow[u][v]=1;
}
floyd(n);
int cnt1,cnt2;
cnt1=0;
for(int i=1;i<=n;i++)
{
cnt2=0;
for(int j=1;j<=n;j++)
{
if(cow[i][j]||cow[j][i])
cnt2++;
}
if(cnt2==n-1)
cnt1++;
}
printf("%d\n",cnt1);
}
return 0;
}

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