hdu 1856 More is better

More is better

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 327680/102400 K (Java/Others)
Total Submission(s): 14656    Accepted Submission(s): 5390

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

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.

 

题意： 给出两两关系， 求最大集合的元素个数。

思路：初始化每个集合的cnt为1，合并过程中合并cnt，并更新最大cnt

换了种写法，p初始化为-1，find_set 时候判断是否是-1，如果是，说明是根节点。

#include<cstring>
#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
const int maxn=10000000+5;
int p[maxn], cnt[maxn], max_cnt;
void make_set()
{
    memset(p, -1, sizeof(p));
    for(int i=0;i<maxn;i++) cnt[i]=1;
    max_cnt=1;
}

int find_set(int x)
{
    return p[x]==-1 ? x : p[x]=find_set(p[x]);
    //return p[x]==x ? x : p[x]=find_set(p[x]);
}

void union_set(int x, int y)
{
    int fx=find_set(x), fy=find_set(y);
    if(fx==fy) return;
    p[fx]=fy;
    cnt[fy]+=cnt[fx];
    max_cnt=max(max_cnt, cnt[fy]);
}

int main()
{
    int n, x, y;
    while(scanf("%d", &n)!=EOF) {
        make_set();
        for(int i=0;i<n;i++)
        {
            scanf("%d%d", &x, &y); union_set(x, y);
        }
        printf("%d\n", max_cnt);
    }
    return 0;
}