F - More is better-优快云博客

王先生希望邀请尽可能多的男孩参与项目，但必须确保他们之间存在直接或间接的友谊关系。通过输入朋友对数及其关系，本博客指导如何选择最大的团队规模。

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题意：求几个集合中最大的（包含子最多）

Crawling in process... Crawling failed Time Limit:1000MS Memory Limit:102400KB 64bit IO Format:%I64d & %I64u

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

1 2

3 4

5 6

1 6

1 2

3 4

5 6

7 8

Sample Output

思路

并查集，求num[i]最大值

代码

#include <stdio.h>
int pre[100007];
int num[1000007];
int find(int h)
{
    if(pre[h]==h)
    return h;
    else
    {
        pre[h]=find(pre[h]);
        return pre[h];
    }
}
void merge(int u,int v)
{
    int t1,t2;
    t1=find(u);
    t2=find(v);
    if(t1!=t2)
    {
        if(num[t1]>num[t2])
        {
            pre[t2]=t1;
            num[t1]+=num[t2];
        }
        else
        {
             pre[t1]=t2;
             num[t2]+=num[t1];
        }
 
    }
}
int main()
{
    int n,i,a,b;
    while(scanf("%d",&n)!=EOF)
    {
        for(i=1;i<100007;i++)
        {
            pre[i]=i;
            num[i]=1;
        }
        while(n--)
        {
            scanf("%d%d",&a,&b);
            merge(a,b);
        }
        int max=num[1];
        for(i=1;i<100007;i++)
        {
            if(num[i]>=max)
                max=num[i];
        }
        printf("%d\n",max);
    }
    return 0;
}