HDU 1856 More is better

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本文通过一个具体的编程问题，介绍了并查集算法的应用场景及其实现细节。问题背景为帮助王先生选择最多的男孩来完成项目，确保留下的男孩直接或间接都是朋友关系。文章详细展示了如何使用并查集算法进行高效的数据管理和查询，最终找到最大数量的符合条件的男孩。

More is better

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 327680/102400 K (Java/Others)
Total Submission(s): 8268 Accepted Submission(s): 3088

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

继续裸并查集，模板搞起~

#include <stdlib.h> 
#include <iostream>
#include <algorithm>
using namespace std;
const int M = 100005;
int fam[M],member[M];
int cas,n,a,b;

void ufset( )
{
     for( int i=0 ; i<M ; i++ )
          member[i] = 1 , fam[i] = i;     
}

int find( int x )
{
    return x == fam[x] ? x : fam[x] = find( fam[x] );    
}

void merge( int a , int b )
{
     int x,y;
     x = find( a );
     y = find( b );
     if( x!=y ){
         fam[x] = y;
         member[y] += member[x];    
     }     
}

int main()
{
    while( scanf("%d",&n) == 1 ){
           ufset( ); 
           int big = 1;
           for( int i=0 ; i<n ; i++ ){
                scanf("%d%d",&a,&b);
                if( a > big || b >big )
                    big = max(a,b);
                merge( a,b );
           }
           int Max = 0;
           for( int i=1 ; i<=big ; i++ )
                if( member[i] > Max )
                    Max = member[i];
           printf("%d\n",Max);
    }        
    return 0;
}