M. Frequent Subsets Problem - 状态压缩-2017 ACM-ICPC 亚洲区（南宁赛区）网络赛

The frequent subset problem is defined as follows. Suppose $U$={1, 2,$\dots$,N} is the universe, and S1S_{1}S​1​​,S2S_{2}S​2​​,$\dots$,SMS_{M}S​M​​ are $M$ sets over $U$. Given a positive constant $\alpha$,$0<\alpha \le 1$, a subset $B$ ($B\ne 0$) is α-frequent if it is contained in at least $\alpha M$ sets of S1S_{1}S​1​​,S2S_{2}S​2​​,$\dots$,SMS_{M}S​M​​, i.e. ∣{i:B⊆Si}∣≥αM\left | \left \{ i:B\subseteq S_{i} \right \} \right | \geq \alpha M∣{ i:B⊆S​i​​}∣≥αM. The frequent subset problem is to find all the subsets that are α-frequent. For example, let$U=\{1,2,3,4,5\}$,$M=3$,$\alpha =0.5$, and S1={1,5}S_{1}=\{1, 5\}S​1​​={ 1,5},S2={1,2,5}S_{2}=\{1,2,5\}S​2​​={ 1,2,5},S3={1,3,4}S_{3}=\{1,3,4\}S​3​​={ 1,3,4}. Then there are $3$ α-frequent subsets of $U$, which are $\{1\}$,$\{5\}$ and $\{1,5\}$.

Input Format

The first line contains two numbers $N$ and $\alpha$, where $N$ is a positive integers, and $\alpha$ is a floating-point number between 0 and 1. Each of the subsequent lines contains a set which consists of a sequence of positive integers separated by blanks, i.e., line$i+1$ contains SiS_{i}S​i​​,$1\le i\le M$ . Your program should be able to handle $N$ up to $20$ and $M$ up to $50$.

Output Format

The number of $\alpha$-frequent subsets.

样例输入

15 0.4
1 8 14 4 13 2
3 7 11 6
10 8 4 2
9 3 12 7 15 2
8 3 2 4 5

样例输出

11

/*
题意：
给你一个n表示集合U是从1~n，一个频率0<α<=1
M个集合，M不是给你的要自己统计，最后让你找这样的子集：
在M个集合中出现次数 >= M*α的

题解：
枚举子集，看其在M个集合中出现的次数

优化：
1.如何存这个M个集合？
用二进制的思想，第几位表示数值为几的数在集合中存在
例如用 0011010 表示1、3、4属于集合Si
那么这个数x存在就可以表示为 1<
#include 
#include 
#include 
#include 
using namespace std;
typedef long long ll;
const int maxn=1e6+10;
long long S[55];
int cnt[30];
int main()
{
    string s;
    int n;
    double a;

    cin>>n>>a;
    getchar();
    int m = 0;
    while(getline(cin,s)){

        stringstream ss;
        ss << s; ///从s读进ss
        int tmp;
        while(ss>>tmp){ ///从ss依次提取
           S[m] += (1 << tmp);
           cnt[tmp] ++;
        }
        m++;
    }

   int minnum;
    if(a*m - int(a*m) > 0.0000001) minnum = ceil(a*m);
        else minnum = a*m;

    long long tempans = 0;
    for(int i = 0;i<30;++i){
        if(cnt[i]>=minnum) tempans += (1<= minnum) {
                ans++;break;
            }
        }
    }
    printf("%lld\n",ans);


    return 0;
}