解题思路：给出n对点的关系，求构成多少个环，如果对于点x和点y,它们本身就有一堵墙，即为它们本身就相连，如果find(x)=find(y)，说明它们的根节点相同，它们之间肯定有直接或间接的相连，即形成环

样例的示意图

共3个环

Ice_cream‘s world I

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 642    Accepted Submission(s): 371

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>
 int pre[10005],a[10005];
 int find( int root)
{
	if(root!=pre[root])
	pre[root]=find(pre[root]);
	return pre[root];
}
void unionroot( int root1,int root2)
{
	 int x,y;
	x=find(root1);
	y=find(root2);
	if(x!=y)
	pre[x]=y;
}

int main()
{
	int root1,root2,x,y,i,n,m,tmp;
	while(scanf("%d %d",&n,&m)!=EOF)
	{
		tmp=0;
		for(i=0;i<=10005;i++)
		pre[i]=i;
		while(m--)
		{
			scanf("%d %d",&root1,&root2);
			x=find(root1);
			y=find(root2);
			if(x==y)
			tmp++;
			unionroot(x,y);
		}

		printf("%d\n",tmp);

	}
}

HDU 2120 Ice_cream's world I【并查集】