HDU1856More is better

最新推荐文章于 2020-03-15 22:05:55 发布

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最新推荐文章于 2020-03-15 22:05:55 发布

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分类专栏：并查集

本文链接：https://blog.youkuaiyun.com/xinxiaxindong/article/details/76603006

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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数量问题。该问题要求在给定直接好友对的情况下，找到能够形成的最大好友群组的人数。通过使用并查集的数据结构，我们可以有效地将直接或间接的好友关系进行合并，并最终确定最大的群组规模。

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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

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

题意

题解：

问一个根结点最多绑定几个节点 0输出1

AC代码

#include<cstdio>
#include<cstring>
#include<stack>
#include <set>
#include <queue>
#include <vector>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long LL;
const int mod = 1e9+7;
int par[10000005];
int rnk[10000005];
int n;
int ans;
void init(){
    for (int i = 1;i <= 10000005; ++i){
        par[i] = i;
        rnk[i] = 1;
    }
}
int fd(int x){
    if (x == par[x]) return x;
    return par[x]=fd(par[x]);
}
void ut(int x,int y){
    int tx = fd(x);
    int ty = fd(y);
    if (tx!=ty){
       par[tx] = ty;
       rnk[ty]+=rnk[tx];
       ans = max(ans,rnk[ty]);
    }
}
int main()
{
    int n;
    while (scanf("%d",&n)!=EOF){
        init();
        if (n==0)
        {
            printf("1\n");
            continue;
        }
        ans = 0;
        int x,y;
        while (n--){
            scanf("%d%d",&x,&y);
            ut(x,y);
        }
        printf("%d\n",ans);
    }
    return 0;
}