HDU1856：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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最新推荐文章于 2021-10-25 19:11:05 发布

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文章标签： java

原文链接：http://www.cnblogs.com/junior19/p/6730018.html

本文详细介绍了并查集算法的应用场景及其实现方式，通过解决一个具体的问题——如何找到最多能够保持互相为朋友关系的学生数量，深入浅出地解析了并查集算法的工作原理。

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More is better

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 327680/102400 K (Java/Others)
Total Submission(s): 25930 Accepted Submission(s): 9308

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

   4 1 2 3 4 5 6 1 6 4 1 2 3 4 5 6 7 8
  

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

并查集入门

# include <stdio.h>
# include <algorithm>
# define MAXN 10000000
using namespace std;
int pre[MAXN+1], num[MAXN+1];

void init()
{
    for(int i=1; i<=MAXN; ++i)
    {
        pre[i] = i;
        num[i] = 1;
    }
}

int find(int x)
{
    if(x != pre[x])
        pre[x] = find(pre[x]);
    return pre[x];
}

int main()
{
    int n, x, y, imax;
    while(~scanf("%d",&n))
    {
        if(n==0)
        {
            puts("1");
            continue;
        }
        init();
        imax = -1;
        while(n--)
        {
            scanf("%d%d",&x,&y);
            int px = find(x);
            int py = find(y);
            if(px != py)
            {
                pre[px] = py;
                num[py] += num[px];
                imax = max(imax, num[py]);
            }
        }
        printf("%d\n",imax);
    }
    return 0;
}

转载于:https://www.cnblogs.com/junior19/p/6730018.html