杭电 1856 more is better

Problem Description

Mr Wang wants some boys to help him with a project. Because the project is rather complex, the more boys come, the better it will be. Of course there are certain requirements.

Mr Wang selected a room big enough to hold the boys. The boy who are not been chosen has to leave the room immediately. There are 10000000 boys in the room numbered from 1 to 10000000 at the very beginning. After Mr Wang's selection any two of them who are still in this room should be friends (direct or indirect), or there is only one boy left. Given all the direct friend-pairs, you should decide the best way.

 

Input

The first line of the input contains an integer n (0 ≤ n ≤ 100 000) - the number of direct friend-pairs. The following n lines each contains a pair of numbers A and B separated by a single space that suggests A and B are direct friends. (A ≠ B, 1 ≤ A, B ≤ 10000000)

 

Output

The output in one line contains exactly one integer equals to the maximum number of boys Mr Wang may keep.

 

Sample Input

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

 

Sample Output

4 2

#include
#define max 10000000
int per[max+100];
int num[max+100];
int find(int x)//找根节点 
{
	int r;
	r=x;
	while(r!=per[r])
	{
		r=per[r];
	}
	per[x]=r;
	return r;
}
void join(int x ,int y)//连接两根节点 
{
	int fx=find(x);
	int fy=find(y);
	if(fx!=fy)
	{
		per[fx]=fy;
		num[fy]+=num[fx];//当前根节点的元素
	}	
}
int main()
{
	int a,b,m,n,i;
	int t;
	while(~scanf("%d",&t))
	{
		if(t==0)
		{
			printf("1\n");
			continue;
		}
		for(i=1;i<=max;++i)//初始化 
		{
			per[i]=i;
			num[i]=1;		//每个节点只有一个元素，自己本身 
		}	
		n=-1;//为了省时间，	
		while(t--)
		{
			scanf("%d%d",&a,&b);
			if(a>n)
			{
				n=a;
			}
			if(b>n)
			{
				n=b;
			}
			join(a,b);
		}	
		int MAX=0;	
		for(i=1;i<=n;++i)//在这用上 
		{
			if(MAX