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 ≤ 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
题目给出了m对的相对关系，求有多少个排名是确定的。
使用floyed求一下传递闭包。如果这个点和其余的关系都是确定的，那么这个点的排名就是确定的。
#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
#define inf 0x3f3f3f3f
int map1[5000][5000];
int main()
{
int k,i,j,n,m;
while(~scanf("%d%d",&n,&m))
{
int a,b;
for(i=1; i<=m; i++)
{
scanf("%d%d",&a,&b);
map1[a][b]=1;
}
for(k=1; k<=n; k++)
for(i=1; i<=n; i++)
for(j=1; j<=n; j++)
if(map1[i][k]==1&&map1[k][j]==1)
map1[i][j]=1;
int sum=0;
for(i=1; i<=n; i++)
{
for(j=1; j<=n; j++)
{
if(i==j)
continue;
if(map1[i][j]==0&&map1[j][i]==0)
break;
}
if(j>n)
sum++;
}
printf("%d
",sum);
}
return 0;
}