思路:一个简单的传递闭包
#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
const int maxn = 105;
int g[maxn][maxn];
int main()
{
int n,m;
while(scanf("%d%d",&n,&m)==2)
{
memset(g,0,sizeof(g));
for(int i = 1;i<=m;i++)
{
int u,v;
scanf("%d%d",&u,&v);
g[u][v]=1;
}
for(int k = 1;k<=n;k++)
for(int i = 1;i<=n;i++)
for(int j = 1;j<=n;j++)
if(g[i][k] && g[k][j])
g[i][j]=1;
int ans = 0;
for(int i = 1;i<=n;i++)
{
int flag = 1;
for(int j = 1;j<=n;j++)
{
if(i==j)continue;
if(!g[i][j] && !g[j][i])
{
flag = 0;
break;
}
}
if(flag)
ans++;
}
printf("%d\n",ans);
}
}
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