| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 7567 | Accepted: 4206 |
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());
}
}
原文地址：http://blog.csdn.net/u010372095/article/details/45201653
