HDOJ 1856 More is better (带权并查集)

最新推荐文章于 2018-11-27 20:45:48 发布

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本文介绍了一个算法问题，即如何从大量的男孩中选出最多的人数，使得任意两人直接或间接为朋友。通过并查集实现，文章提供了完整的代码实现及解析。

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More is better

Time Limit: 5000/1000 MS(Java/Others) Memory Limit: 327680/102400 K(Java/Others)
Total Submission(s): 26824 Accepted Submission(s): 9607

Problem Description

Mr Wang wants some boys tohelp him with a project. Because the project is rather complex, the moreboys 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 beenchosen has to leave the room immediately. There are 10000000 boys in the roomnumbered from 1 to 10000000 at the very beginning. After Mr Wang's selectionany two of them who are still in this room should be friends (direct orindirect), 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 inputcontains 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 asingle space that suggests A and B are direct friends. (A ≠ B, 1 ≤ A, B ≤10000000)

Output

The output in one linecontains exactly one integer equals to the maximum number of boys Mr Wang maykeep.

Sample Input

1 2

3 4

5 6

1 6

1 2

3 4

5 6

7 8

Sample Output

Hint

A and B are friends(direct orindirect), B and C are friends(direct or indirect),

then A and C are alsofriends(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.

#include <bits/stdc++.h>
using namespace std;
#define mst(a,b) memset((a),(b),sizeof(a))
#define f(i,a,b) for(int i=(a);i<=(b);++i)
const int maxn = 10000005;
const int mod = 200907;
#define ll long long
#define rush() int t;scanf("%d",&t);while(t--)
int n,ans,pre[maxn],num[maxn];
void init()
{
    f(i,1,maxn)
    {
        pre[i]=i;
        num[i]=1;
    }
    ans=1;
}
int find(int x)
{
    int t,r=x;
    while(pre[x]!=x)
    {
        x=pre[x];
    }
    while(r!=x)
    {
        t=pre[r];
        pre[r]=x;
        r=t;
    }
    return x;
}
void join(int a,int b)
{
    int A=find(a);
    int B=find(b);
    if(A!=B)
    {
        pre[A]=B;
        num[B]+=num[A];
        if(num[B]>ans)
            ans=num[B];
    }
}
int main()
{
    int x,y;
    while(~scanf("%d",&n))
    {
        init();
        while(n--)
        {
            scanf("%d%d",&x,&y);
            join(x,y);
        }
        printf("%d\n",ans);
    }
    return 0;
}