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本文介绍了一个算法问题，即如何计算在特定条件下最大的好友群数量。问题中，王先生希望找到尽可能多的男孩帮助完成一个项目，并规定了他们之间的友谊关系。通过输入直接朋友对的数据，算法实现了寻找最大群体的过程。

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Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 327680/102400 K (Java/Others)
Total Submission(s): 11653 Accepted Submission(s): 4319

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.

这个题要注意输入m=0时，结果应该输出1；且需路径压缩。。否则会超时的！

#include<stdio.h>
#include<string.h>
#include<malloc.h>
#include<stdlib.h>
int bin[1100000],num[1100000];
int u[1100000],w[1100000];
int findx(int x)
{
    int i,j;
    int r=x;
    while(bin[r]!=r)
    {
        r=bin[r];
    }

    i=x;
    while(i!=r)
    {
        j=bin[i];
        bin[i]=r;
        i=j;
    }
    return r;
}
void merge(int x,int y)
{
    int fx,fy;
    fx=findx(x);
    fy=findx(y);
    if(fx!=fy)
        bin[fx]=fy;
}
int main()
{
    int n,m,i,j;
    while(~scanf("%d",&m))
    {
        if(m==0)
            printf("1\n");
        else
        {
            int maxx=0;
            for(i=0; i<m; i++)
            {
                scanf("%d%d",&u[i],&w[i]);
                int e=u[i]>w[i]?u[i]:w[i];
                if(e>maxx)
                    maxx=e;
            }
            for(i=1; i<=maxx; i++)
            {
                bin[i]=i;
            }
            for(i=0; i<m; i++)
            {
                merge(u[i],w[i]);
            }
            int max=0;
            memset(num,0,sizeof(num));
            for(i=1; i<=maxx; i++)
            {
                num[findx(i)]++;
                if(num[findx(i)]>max)
                    max=num[findx(i)];
            }
            printf("%d\n",max);
        }
    }
    return 0;
}