HDU1856 More is better 解题报告

More is better

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 327680/102400 K (Java/Others)
Total Submission(s): 10251    Accepted Submission(s): 3778

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.

 

Author

lxlcrystal@TJU

 

Source

HDU 2007 Programming Contest - Final

 

Recommend

lcy

#include 
using namespace std;
#define N 10000000
int father[N],cnn[N];//cnn存的是这个点包含的子集个数
int Find(int u)
{
	if(father[u]!=u)
		return father[u]=Find(father[u]);
	else
		return father[u];
}
void Union(int a,int b)
{
	a=Find(a);
	b=Find(b);
	if (a!=b)
	{
		if(cnn[a]>cnn[b])//如果a包含的子集更多
		{
			father[b]=a;//b归属于a即b的父亲节点为a
			cnn[a]+=cnn[b];//b的子集也归a
		}
		else
		{
			father[a]=b;
			cnn[b]+=cnn[a];
		}
	}
	return;
}
int main()
{
    int n;
	while(~scanf("%d",&n))
	{
		int i;
		for(i=1;i<=N;i++)//每次都是N个
		{
			father[i]=i;
			cnn[i]=1;//初始每个点的子集都只有自己
		}
		while(n--)
		{
			int a,b;
			scanf("%d%d",&a,&b);
			Union(a,b);
		}
		int mma=0;
		for(i=1;i<=N;i++)
		{
			if(cnn[i]>mma)
				mma=cnn[i];//找出子集个数最多的点
		}
		printf("%d\n",mma);//输出最多子集个数
	}
	return 0;
}