H - Cow Contest

题目链接：

题目：

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.

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

5 54 34 23 21 22 5

Sample Output

2 题意：给出n,m，n是n头牛，m是m对关系，a,b，在前的即为a打败了b，一开始没看懂让我们求啥，然后百度发现用到了传递闭包，有向图的传递背包算法 其中包括了floyed传递背包算法，就是1跟2有关系，2跟3有关系，1也就跟3有关系，算这些牛中能确立多少等级关系，一旦一头牛和其他牛都存在胜负关系， 那么就确立一个等级关系，最后for循环遍历一下就可算出几个等级关系

//// Created by hanyu on 2019/7/20.//#include 
      
       #include 
       
        #include 
        
         #include 
         
          using namespace std;const int maxn=105;#define MAX 0x3f3f3f3fint road[maxn][maxn];int main(){    int n,m;    scanf("%d%d",&n,&m);    int front,to;    memset(road,MAX,sizeof(road));    for(int i=1;i<=n;i++)        road[i][i]=0;//自己和自己不成立关系    for(int i=1;i<=m;i++)    {        scanf("%d%d",&front,&to);        road[front][to]=1;//单向图        road[to][front]=-1;//反向不成立    }    for(int k=1;k<=n;k++)    {        for(int i=1;i<=n;i++)        {            for(int j=1;j<=n;j++)            {                if(road[i][k]==road[k][j]&&road[i][k]!=MAX)                {                    road[i][j]=road[i][k];                }            }        }    }//Floyd 传递闭包算法    int cnt,sum=0;    for(int i=1;i<=n;i++)    {        cnt=0;        for(int j=1;j<=n;j++)        {            if(i!=j&&road[i][j]!=MAX)            {                cnt++;            }//遍历每头牛和其他牛的关系        }        if(cnt==n-1)//如果关系达到n-1，即和自身之外的其他牛都确定了关系，则可确定等级            sum++;    }    printf("%d\n",sum);    return 0;}