hdu 1856 More is better 基础并查集★

最新推荐文章于 2021-10-25 19:11:05 发布

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最新推荐文章于 2021-10-25 19:11:05 发布

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本文介绍了一个关于最大连通量的问题，通过描述王先生希望通过选取一些男孩来帮助完成一个复杂的项目，引出了如何通过给定的直接朋友对来确定最多可以保留多少个男孩的方法。使用并查集算法进行求解，最终输出王先生可以保留的最大男孩数量。

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More is better

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 327680/102400 K (Java/Others)
Total Submission(s): 29398 Accepted Submission(s): 10436

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

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最大连通量

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int pre[1000003];
int num[1000003];
int res=0;
void init()
{
    for(int i=0;i<=1000000;i++)
        pre[i]=i,num[i]=1;
}
int Find(int x)
{
    if(x!=pre[x])
        return pre[x]=Find(pre[x]);
    return pre[x];
}
void Merge(int x,int y)
{
    int X=Find(x);
    int Y=Find(y);
    if(X!=Y)
    {
        pre[X]=Y;
        num[Y]+=num[X];
        res=max(res,num[Y]);
    }
}

int main()
{
    int n;
    while (~scanf("%d", &n))
    {
        init();
        res=1;
        for(int i = 0; i < n; i++)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            Merge(x,y);
        }
        printf("%d\n",res);
    }
}