hdu2120 Ice_cream's world I(検索ループの個数を調べる)

Ice_cream's world I
Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 887    Accepted Submission(s): 517
Problem Description
ice_cream's world is a rich country, it has many fertile lands. Today, the queen of ice_cream wants award land to diligent ACMers. So there are some watchtowers are set up, and wall between watchtowers be build, in order to partition the ice_cream’s world. But how many ACMers at most can be awarded by the queen is a big problem. One wall-surrounded land must be given to only one ACMer and no walls are crossed, if you can help the queen solve this problem, you will be get a land.
 
Input
In the case, first two integers N, M (N<=1000, M<=10000) is represent the number of watchtower and the number of wall. The watchtower numbered from 0 to N-1. Next following M lines, every line contain two integers A, B mean between A and B has a wall(A and B are distinct). Terminate by end of file.
 
Output
Output the maximum number of ACMers who will be awarded.
One answer one line.
 
Sample Input

   
   
   
   

    
    
    
    8 10
0 1
1 2
1 3
2 4
3 4
0 5
5 6
6 7
3 6
4 7
   
   
   
   

 
Sample Output

   
   
   
   

    
    
    
    3
   
   
   
   

 
Author
Wiskey
 
Source
HDU 2007-10 Programming Contest_WarmUp
 
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この問題の意味は1つの図の中にいくつかの環があることを探すことです.
私のすべての知識があまりにも死んだせいだ.ああ、最初は思いもよらなかった.のこのような問題をもういくつかほしい.
拾わせてください.の

#include <stdio.h>
int fa[1005];
int find(int x)
{
	if(fa[x]!=x) fa[x]=find(fa[x]);
	return fa[x];
}
bool comb(int a,int b)
{
	a=find(a);
	b=find(b);
	if(a==b)//           
	return true;
	else
	{
		fa[a]=b;
		return false;
	}
}
void init(int n)
{
	for(int i=0;i<n;i++)
	fa[i]=i;
}
int main()
{
	int n,k;
	while(scanf("%d %d",&n,&k)!=EOF)
	{
		init(n);
		int sum=0;
		for(int i=0;i<k;i++)
		{
			int a,b;
			scanf("%d %d",&a,&b);
			if(comb(a,b))
			sum++;
		}
		printf("%d
",sum);
	}
	return 0;
}