hdu 1856 More is better(并查集)

并查集解决好友关系问题

最新推荐文章于 2021-10-25 19:11:05 发布

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本文介绍了一个基于并查集算法的好友关系处理问题，旨在通过算法将直接或间接成为朋友的人群进行分组，以求得最大群体数量。问题设定在一个包含一千万名成员的房间里，通过输入的直接好友关系对，利用并查集算法合并相关联的成员，最终输出最大可能保留的成员数。

More is better

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 327680/102400 K (Java/Others)
Total Submission(s): 8057 Accepted Submission(s): 3001

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

思路：并查集，简单把是朋友的都合并就好了。注意0的情况。

#include<iostream>
#include<cstring>
using namespace std;
const int mm=1e7+9;
int root[mm],ran[mm];
int n,a,b,ans;
int look(int x)
{
  if(x^root[x])root[x]=look(root[x]);
  return root[x];
}
void data_set()
{
  for(int i=0;i<mm;i++)
    root[i]=i,ran[i]=1;
}
void uni(int x,int y)
{
  x=look(x);y=look(y);
  if(x==y)return;
  if(ran[x]>ran[y])root[y]=x,ran[x]+=ran[y];
  else root[x]=y,ran[y]+=ran[x];
  if(ans<ran[x])ans=ran[x];
  if(ans<ran[y])ans=ran[y];
}
int main()
{
  while(cin>>n)
  { data_set();ans=0;
    for(int i=0;i<n;i++)
    {
      cin>>a>>b;
      if(a^b)
      uni(a,b);
    }
    if(n==0)ans=1;
    cout<<ans<<"\n";

  }
}