HDU-1856-More is better【并查集】

最新推荐文章于 2021-10-25 19:11:05 发布

Somethingwll

最新推荐文章于 2021-10-25 19:11:05 发布

阅读量428

点赞数

CC 4.0 BY-SA版权

分类专栏：大菜鸟小历程并查集

本文链接：https://blog.youkuaiyun.com/wang907553141/article/details/52087235

大菜鸟小历程同时被 2 个专栏收录

155 篇文章

订阅专栏

并查集

13 篇文章

订阅专栏

本文介绍了一个算法问题，即如何从一系列直接好友关系中找到最大的间接好友群体。通过使用并查集的方法，有效地解决了该问题，并提供了完整的代码实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

More is better

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 327680/102400 K (Java/Others)

Total Submission(s): 24270 Accepted Submission(s): 8711

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.

找出某个集合的元素个数，且保证此集合中的元素个数是所有集合元素个数中最大的

#include<cstdio>
#include<algorithm>
using namespace std;
const int Q=1e7;
int a,b,fa[Q+10],cnt[Q+10];
int find(int x)
{
	if(x==fa[x])	return x;
	return fa[x]=find(fa[x]);
}
void Union(int x,int y)
{
	int nx=find(x);
	int ny=find(y);
	if(nx!=ny)
	{
		fa[ny]=nx;
		cnt[nx]+=cnt[ny];  // 合并两个集合的元素个数，注意合并方向 
	}
}
int main()
{
	int n;
	while(~scanf("%d",&n))
	{
	/*	if(n==0)	// 这一段加不加无所谓了 
		{
			printf("1\n");
			continue;
		}*/
		for(int i=1;i<=Q;i++)
		{
			fa[i]=i;
			cnt[i]=1;  // 刚开始 Q 个集合，且每个集合中只有一个元素 
		}
		int num_max=0;
		while(n--)
		{
			scanf("%d %d",&a,&b);
			num_max=max(num_max,max(a,b));
			Union(a,b);
		}
		int ans=0;
		for(int i=1;i<=Q;i++)
		{
			if(ans<cnt[i])
				ans=cnt[i];
		}
		printf("%d\n",ans);
	}
	return 0;
}