POJ 3660Cow Contest（並查集+拓撲排序）

POJ 3660Cow Contest（並查集+拓撲排序）

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

題意：有n個牛，編號1~n。現給出m條關系：A B ，說明A比B厲害。現在問有多少個牛能被唯一確定（即這頭牛與n-1頭牛的關系是唯一確定的）。

解題：一頭牛如果能被唯一確定，那麼所有的點一定是一個連通塊，那麼就可能用到並查集 來判斷。再接下來就是拓撲排序了。具體看代碼。

#include
#include

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(l0)
    {
        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)//說明所有的點不是在一個連通塊內，所有的點都不能被確定
            printf("0\n");
        else
            printf("%d\n",tope());
    }
}