HDU 1856（并查集）

More is better

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 327680/102400 K (Java/Others)
Total Submission(s): 1576    Accepted Submission(s): 590

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


    
    
    
    
 
     
 
      Hint 
     
 A and B are friends(direct or indirect), B and C are friends(direct or indirect), then A and C are also friends(indirect). In the first sample {1,2,5,6} is the result. In the second sample {1,2},{3,4},{5,6},{7,8} are four kinds of answers. 
    
   
   
   
   
   
   
   
   

    
    
    
     
   
   
   
   
   
   
   
   

    
    
    #include <iostream>
using namespace std;
#define N 100005

struct set
{
	int parent;  //记录父节点
	int rank;    //记录集合的节点数
}elem[N];

int MAX;

void init()
{
	int i;
	for(i=0;i<=N;i++)
	{
		elem[i].parent=i;
		elem[i].rank=1;
	}
}

int Find(int x)
{
	int root,temp;
	temp=x;
	while(x!=elem[x].parent)    //寻找根节点
		x=elem[x].parent;
	root=x;
	x=temp;
	while (x!=elem[x].parent)   //压缩路径，全部赋值为根节点的值
	{
		temp=elem[x].parent;
		elem[x].parent=root;
		x=temp;
	}
	return root;
}

void Union(int a,int b)   //合并两个集合
{
	int x,y;
	x=Find(a);
	y=Find(b);
	if(elem[x].rank>=elem[y].rank)
	{
		elem[y].parent=elem[x].parent;
		elem[x].rank+=elem[y].rank;
		if(MAX<elem[x].rank)
			MAX=elem[x].rank;
	}
	else
	{
		elem[x].parent=elem[y].parent;
		elem[y].rank+=elem[x].rank;
		if(MAX<elem[y].rank)
			MAX=elem[y].rank;
	}
}

int main()
{
	int n;  //有关系的对数
	int a,b,x,y;
	while (scanf("%d",&n)!=EOF)
	{
		init();
		MAX=-1;
		while (n--)
		{
			scanf("%d%d",&a,&b);
			x=Find(a);
			y=Find(b);
			if(x!=y)    
				Union(a,b);//a和b不是一个集合的，合并这两个集合
		}
		if(MAX!=-1)
			printf("%d/n",MAX);   //输出最大集合的节点数
		else
			printf("1/n");   //此时n=0，证明都是没关系的人
	}
	return 0;
}