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 5 4 3 4 2 3 2 1 2 2 5
Sample Output
2
floyd判联通 也有dfs的写法 map[i][j]之间是联通的那就表明他们之间确立了强弱关系
#include<iostream>
#include<cstring>
using namespace std;
int map[120][120];
int n,m;
void floyd()
{
for(int k=1;k<=n;k++)
{
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
if(map[i][k]==1&&map[k][j]==1) //如果联通就是可以 确定排名
map[i][j]=1;
}
}
}
}
int main()
{
ios::sync_with_stdio(false);
while(cin>>n>>m)
{
memset(map,0,sizeof(map));
int ans=0;
if(!n||!m)
{
break;
}
for(int i=1;i<=m;i++)
{
int s,e;
cin>>s>>e;
map[s][e]=1;
}
floyd();
for(int i=1;i<=n;i++)
{
int cnt=0;
for(int j=1;j<=n;j++)
{
if(map[i][j]==1||map[j][i]==1)
{
cnt++;
}
}
ans=ans+((cnt==n-1)?1:0);
}
cout<<ans<<endl;
}
}