HDU-2120-Ice_cream's world I【并查集】

Ice_cream's world I

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1265    Accepted Submission(s): 758

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

题解：求由 0 到 n-1 围成的大区域被墙给分成了几个小区域，上图：

#include<cstdio>
#include<algorithm>
const int Q=1e4;
int n,m,fa[Q+10];
int find(int x)
{
	if(x==fa[x])	return x;
	return fa[x]=find(fa[x]);
}
void Union(int x,int y)
{
	int nx=find(x);
	int ny=find(y);
	if(nx!=ny)
	{
		fa[ny]=nx;
	}
}
int main()
{
	while(~scanf("%d %d",&n,&m))
	{
		for(int i=0;i<n;i++)
		{
			fa[i]=i;
		}
		int a,b,ans=0;
		while(m--)
		{
			scanf("%d %d",&a,&b);
			if(find(a)==find(b))
			{
				ans++;
			}
			Union(a,b);
		}
		printf("%d\n",ans);
	}
	return 0;
}