hdoj 2120 Ice_cream's world I【求成环数】

Ice_cream's world I

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 741    Accepted Submission(s):
429

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

#include<stdio.h>
#include<string.h>
#define MAX 1010
int set[MAX];
int sum=0;
int find(int fa)
{
	int t;
	int ch=fa;
	while(fa!=set[fa])
	fa=set[fa];
	while(ch!=fa)
	{
		t=set[ch];
		set[ch]=fa;
		ch=t;
	}
	return fa;
}
void mix(int x,int y)
{
	int fx,fy;
	fx=find(x);
	fy=find(y);
	if(fx!=fy)
	set[fx]=fy;
	else
	sum++;
}
int main()
{
	int n,m,i,a,b;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		for(i=0;i<=n;i++)
		set[i]=i;
		for(i=0;i<m;i++)
	    {
		    scanf("%d%d",&a,&b);
		    mix(a,b);
	    }
	    printf("%d\n",sum);
	    sum=0;
	}
	return 0;
}