hdu 1856 More is better 并查集（二）

最新推荐文章于 2021-10-25 19:11:05 发布

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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1千万名候选人中选出最多的好友群组成员，以协助完成一项复杂的项目。文章详细解释了并查集的数据结构和操作流程，并提供了完整的C++实现代码。

More is better

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

#include<iostream>
#include<stdio.h>
using namespace std;

int father[10000001],num[10000001];

/*初始化数组*/
void makeset(int x)
{
    for(int i=0;i<=x;i++)
    {
        father[i]=i;
        num[i]=1;
    }
}

int findset(int x)//查
{
    if(x!=father[x])
    {
        father[x]=findset(father[x]);
    }//回溯
    return father[x];
}

void Union(int a,int b)//并
{
int x=findset(a);
int y=findset(b);
      if(x==y)
      {
          return;
      }
      if(num[x]>=num[y])
      {
          father[y]=x;
          num[x]+=num[y];
      }
      else
      {
          father[x]=y;
          num[y]+=num[x];
      }
}

int main()
{ int n;

    while(scanf("%d",&n)!=EOF)
    {
          makeset(10000000);
        for(int i=1;i<=n;i++)
            {
                int first,second;
                scanf("%d%d",&first,&second);
                Union(first,second);
            }
        int max=-99999;
        for(int i=1;i<=10000000;i++)
            {
                if(num[i]>max)
                max=num[i];
            }
        printf("%d\n",max);
    }
    return 0;
}