| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions:14435 | Accepted: 8083 |
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
Source
题目大意：题意很简单，就是给牛确定排名关系，给出n条牛，以及m个比赛状况：A:B谁在前谁就赢，问最后有多少头牛的排名关系是确定的
思路：上来我是没有思路的，然后想了一会，突然想到把赢的一方都用1初始化，然后输的都用-1初始化，然还是没用，最后百度了代码，发现用到什么传递闭包…………
传递闭包：
所谓传递性，可以这样理解：对于一个节点i，如果j能到i，i能到k，那么j就能到k。求传递闭包，就是把图中所有满足这样传递性的节点都弄出来，计算完成后，我们也就知道任意两个节点之间是否相连。
传递闭包的计算过程一般可以用Warshell算法描述：
For 每个节点i Do
For 每个节点j Do
If j能到i Then
For 每个节点k Do
a[j, k] := a[j, k] Or ( a[j, i] And a[ i, k] )
其中a数组为布尔数组，用来描述两个节点是否相连，可以看做一个无权图的邻接矩阵。可以看到，算法过程跟Floyd很相似，三重循环，枚举每个中间节点。只不过传递闭包只需要求出两个节点是否相连，而不用求其间的最短路径长。
一头牛的等级，当且仅当它与其它N-1头牛的关系确定时确定，于是我们可以将牛的等级关系看做一张图，然后进行适当的松弛操作，得到任意两点的关系，再对每一头牛进行检查即可代码：
#include<queue>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
using namespace std;
#define ll long long
#define inf 99999999
#define N 111
int e[N][N];
int dis[N];
int book[N];
int m,n;
int main()
{
scanf("%d%d",&n,&m);
memset(e,0,sizeof(e));//初始化为0
int i,j,a,b,c,ans=0,k=0;
for(i=1; i<=m; i++)
{
scanf("%d%d",&a,&b);
e[a][b]=1;//初始化，a赢b
e[b][a]=-1;//b输a;
}
for(k=1; k<=n; k++)
{
for(i=1; i<=n; i++)
{
for(j=1; j<=n; j++)
{
if( e[i][k]==e[k][j]&&(e[i][k]==1||e[i][k] == -1 ) )
e[i][j]=e[i][k];
}
}
}
// for(i=0;i<=n;i++)
// {
// for(j=1;j<=n;j++)
// printf("%10d ",e[i][j]);
// printf("\n");
// }
for(i=1;i<=n;i++)
{
int sum=0;
for(j=1;j<=n;j++)
{
if(e[i][j])
sum++;
}
if(sum==n-1)//与剩下的n-1头牛的胜负关系确定
ans++;
}
printf("%d\n",ans);
return 0;
}