A - Ice_cream's world I HDU - 2120

最新推荐文章于 2020-12-15 10:47:23 发布

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在冰激凌王国，勤劳的ACMer们有机会获得女王颁发的土地奖励。通过设立哨塔和建造城墙，王国被划分为多个区域。挑战在于确定最多能有多少ACMer获得奖励，每个被城墙围住的土地必须单独授予一位ACMer，且城墙不能交叉。解决这个问题将帮助女王公正地分配土地。

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

Sample Output

#include<stdio.h>
int fa[1005];
int findroot(int x)
{
	while(x!=fa[x])
	  x=fa[x];
	return x;
}
int Union(int x,int y)
{
	int nx,ny;
	nx=findroot(x);
	ny=findroot(y);
	if(nx!=ny)
	{
		fa[nx]=ny;
		return 1;
	}
	return 0;
}
int main()
{
	int n,m,i;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		for(i=0;i<1005;i++)
		   fa[i]=i;
		int a,b,flag=0;
		for(i=0;i<m;i++)
		{
			scanf("%d%d",&a,&b);
			if(!Union(a,b))
			  flag++;
		}
		printf("%d\n",flag);
	}
	return 0;
}