hdu_1856_More is better_201403091720

More is better

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 327680/102400 K (Java/Others)
Total Submission(s): 11893    Accepted Submission(s): 4402

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.

 

 

Author

lxlcrystal@TJU

 

 

Source

HDU 2007 Programming Contest - Final

 

题意：王老师要找一些男生帮助他完成一项工程。要求最后挑选出的男生之间都是朋友关系，可以说直接的，也可以是间接地。问最多可以挑选出几个男生（最少挑一个)。

         这是一个有关并查集的题目。只不过不是求有几个集合，而是求每个集合中元素的个数，进而求出个数的最大值。和求集合个数的方法差不多，只需要在合并两个集合时处理一下，让这两个集合的元素个数也合并一下就行了。接下来只需要找出最大值即可。要注意的一个地方就是：当n=0时，要输出1。

 

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#define MAX 10000010

using namespace std;

int pre[MAX],s[MAX];

int find(int x)
{
    int i,t,r;
    r=x;
    while(r!=pre[r])
    r=pre[r];
    while(x!=r)
    {
        i=pre[x];
        pre[x]=r;
        x=i;
    }
    return r;
}
int main()
{
    int i,n;
    while(scanf("%d",&n)!=EOF)
    {
        int i,j,k,a,b,t=0,pa,pb,max=0;
        memset(pre,0,sizeof(pre));
        memset(s,0,sizeof(s));
        if(n==0)
        {
            printf("1\n");
            continue;
        }
        for(i=1;i<MAX;i++)
        {
            pre[i]=i;
            s[i]=1; //开始时数量都为1，根节点为自己 
        }
        for(i=0;i<n;i++)
        {
            scanf("%d %d",&a,&b);
            if(a>t) t=a;
            if(b>t) t=b;
            pa=find(a);pb=find(b);
            //printf("%d %d\n",pa,pb);
            if(pa!=pb)
            {
                pre[pb]=pa;
                s[pa]+=s[pb];   //合并集合中元素个数
            }
        }
        for(i=1;i<=t;i++)
        {
            if(s[i]>max)
            max=s[i];
        }
        printf("%d\n",max);
    }
    return 0;        
}