hdu 1856 More is better（并查集）

题目：​​http://acm.hdu.edu.cn/showproblem.php?pid=1856​​

More is better

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 327680/102400 K (Java/Others)
Total Submission(s): 19461    Accepted Submission(s): 7167

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

分析：并查集应用，需要统计每个集合的大小，所以需要在合并的时候注意更新相应的信息。

#include <iostream>
#include <cstdio>
using namespace std;
const int maxn=10000000;
int father[maxn+1],sum[maxn+1],n; 
int find(int x){
    if(x==father[x]) return x;
    return father[x]=find(father[x]);
}
int main()
{
    //freopen("cin.txt","r",stdin);
    int a,b,i,j;
    while(cin>>n){
        if(n==0){
             puts("1"); //or there is only one boy left
             continue;
        }
        for(i=1;i<=maxn;i++){  //数组越界，发生输出多了少了的怪事儿。。
            father[i]=i;
            sum[i]=1;
        }
        int maxm=0;
        for(i=0;i<n;i++){
            scanf("%d%d",&a,&b);
            maxm=max(max(a,b),maxm);
            int q1=find(a),q2=find(b);
            if(q1!=q2){
                father[q1]=father[q2];  //q1 --> q2
                sum[q2]+=sum[q1];
            }
        }
        int res=0;
        for(i=1;i<=maxm;i++){
            res=max(res,sum[i]);
        }
        printf("%d\n",res);
    }
    return 0;
}