HDU1856 H - More is better(并查集）

最新推荐文章于 2021-10-25 19:11:05 发布

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本文深入探讨了并查集算法的应用场景，特别是在处理大规模社交网络关系中的朋友对问题时，如何通过并查集来确定最大可能保留的个体数量。文章详细介绍了算法实现，包括初始化、查找和合并操作，并通过一个具体示例展示了算法的有效性和效率。

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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)

当N为0时输入结束。
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

这道题是一个并查集的题目，只不过不是问有多少个集合，而是找集合中的元素个数，只需要合并的时候将合并的子节点数目加到父节点上就好了。

#include<bits/stdc++.h>
#define fi first
#define se second
#define FOR(a) for(int i=0;i<a;i++)
#define show(a) cout<<a<<endl;
#define show2(a,b) cout<<a<<" "<<b<<endl;
#define show3(a,b,c) cout<<a<<" "<<b<<" "<<c<<endl;
using namespace std;

int f[10000005],vis[10000005];



int fid(int x)
{
	if(f[x]==x) return x;
	else return f[x]=fid(f[x]);
}

void joint(int x,int y)
{
	int fx=fid(x);
	int fy=fid(y);
	if(fx!=fy)
	{
		f[fx]=fy;
		vis[fy]+=vis[fx];//将子节点上的数目加到父节点上
	}
}
int main()
{
	int n,m,x,y,t;
	while(scanf("%d",&n)!=EOF)

	{


		int sum=1,t=1,flag=0;
		int cnt=0,x,y;

		for(int i=1;i<=10000005;i++) f[i]=i;
		for(int i=1;i<=10000005;i++) vis[i]=1;// 初始为每个都为1
		for(int i=0;i<n;i++)
		{
			scanf("%d%d",&x,&y);

			//show2(fid(x),fid(y));
			t=max(x,t);
			t=max(y,t);
			joint(x,y);

		}
		//cout<<t<<endl;

		for(int i=1;i<=t;i++)
		{

			cnt=max(cnt,vis[i]);

		}
		printf("%d\n",cnt);



	}
}