HDU1856 More is better 解题报告

最新推荐文章于 2020-04-29 14:49:35 发布

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最新推荐文章于 2020-04-29 14:49:35 发布

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分类专栏：并查集文章标签： HDU

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本文介绍了一种通过并查集算法解决的最大好友群组问题，旨在帮助王先生从1千万名编号的男孩中选出最大的互为直接或间接好友的群体。通过输入直接好友对的数据，利用并查集进行数据合并，最终输出最大群组的人数。

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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

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 <iostream>
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;
}