POJ 3660Cow Contest（并查集+拓扑排序）

Cow Contest

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 cowB.

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

题意：有n个牛，编号1~n。现给出m条关系：A B ，说明A比B厉害。现在问有多少个牛能被唯一确定（即这头牛与n-1头牛的关系是唯一确定的）。

解题：一头牛如果能被唯一确定，那么所有的点一定是一个连通块，那么就可能用到并查集 来判断。再接下来就是拓扑排序了。具体看代码。

#include<stdio.h>
#include<string.h>

const int N = 105;

bool mapt[N][N],path[N][N];
int n,in[N],father[N];

void init()
{
    for(int i=1;i<=n;i++)
    {
        father[i]=i; in[i]=0;
        for(int j=1;j<=n;j++)
         mapt[i][j]=path[i][j]=0;
        path[i][i]=1;
    }
}
int findroot(int x)
{
    if(x!=father[x])
        father[x]=findroot(father[x]);
    return father[x];
}
void setroot(int x,int y)
{
    x=findroot(x);
    y=findroot(y);
    father[x]=y;
}

int tope()
{
    int a[N],k=0,l=0,ans=0;
    for(int i=1;i<=n;i++)
    if(in[i]==0)
      a[k++]=i;
    while(l<k)
    {
        int s=a[l++];
        if(l==k)//只有当一个点时，这个点才有可能被确定
        {
            int i;
            for(i=0;i<k;i++)//前方出现的点，看有没有都直接或介接的指向（都有一条有向路可走到s点）
             if(path[s][a[i]]==0)
             break;
            if(i==k)
                ans++;//,printf("%d ",s)//输出可确定的点
        }

        for(int j=1;j<=n;j++)
        if(mapt[s][j])
        {
            in[j]--;
            for(int i=0;i<k;i++)//合并
                path[j][a[i]]|=path[s][a[i]];
            if(in[j]==0)
                a[k++]=j;
        }
    }
    return ans;
}
int main()
{
    int a,b,m;
    while(scanf("%d%d",&n,&m)>0)
    {
        init();
        while(m--)
        {
            scanf("%d%d",&a,&b);
            setroot(a,b);
            if(mapt[a][b]==0)
             mapt[a][b]=1,in[b]++;
        }
        int k=0;
        for(int i=1;i<=n;i++)
            if(father[i]==i)
            k++;
        if(k>1)//说明所有的点不是在一个连通块内，所有的点都不能被确定
            printf("0\n");
        else
            printf("%d\n",tope());
    }
}