HDU 3006 The Number of set (状态压缩)

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本文探讨了一个集合组合问题，即给定多个整数集合，求所有可能组合成的不同新集合的数量。利用状态压缩的方法，通过位运算实现了集合的组合计算。

The Number of set

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1472 Accepted Submission(s): 921

Problem Description

Given you n sets.All positive integers in sets are not less than 1 and not greater than m.If use these sets to combinate the new set,how many different new set you can get.The given sets can not be broken.

Input

There are several cases.For each case,the first line contains two positive integer n and m(1<=n<=100,1<=m<=14).Then the following n lines describe the n sets.These lines each contains k+1 positive integer,the first which is k,then k integers are given. The input is end by EOF.

Output

For each case,the output contain only one integer,the number of the different sets you get.

Sample Input

Sample Output

  
   15
2

Source

2009 Multi-University Training Contest 11 - Host by HRBEU

题意：

给你N个集合，求出N个集合相互组合后最多的集合数。

POINT：

状态压缩：

即把每个数字取与不取用1和0表示，这样得到一排101010即二进制数，之后用各种位运算来实现各种操作。

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <queue>
#include <algorithm>
#include <map>
#include <math.h>
using namespace std;
#define ll long long
int a[(1<<15)+1];
int main()
{
    int n,m;
    while(~scanf("%d %d",&n,&m))
    {
        memset(a,0,sizeof a);
        for(int i=1;i<=n;i++)
        {
            int num;
            scanf("%d",&num);
            int k=0;
            for(int j=1;j<=num;j++)
            {
                int a;
                scanf("%d",&a);
                k+=1<<(a-1);
            }
            a[k]=1;
            for(int j=1;j<=(1<<m);j++)
            {
                if(a[j])
                {
                    a[k|j]=1;//k|j代表把两种集合结合。
                }
            }
        }
        int ans=0;
        for(int i=1;i<=(1<<m);i++)
        {
            if(a[i]) ans++;
        }
        printf("%d\n",ans);
    }
}