M. Frequent Subsets Problem

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 0.41 8 14 4 13 23 7 11 610 8 4 29 3 12 7 15 28 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<=1M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<<x最大的数是20，也就是说最多21位，最大数值是2^21，long long轻松处理于是我们得到了M个longlong型的整数，这些数在计算机中都是以01串存在的2. 如何枚举子集？for(i = 1;i<=U;++i){if(i & Sj)==i,i是U的一个子集}例如i在计算机中二进制表示为101，Sj在计算机表示为1101如果i是Sj的子集那么按位与出来的结果还是i如果结果 != i 说明存在某些位置i是1 Sj是0 即集合i中存在该元素，集合Sj不存在枚举子集 也不过2^21 * 50 应该超不了一秒，不过这里还可以做一个优化3.我们统计所有元素出现的总次数，如果<M*α，那么所求子集中一定不会出现这些元素。于是我们得到了一个新集合tempans ，枚枚举该集合的子集即可即 for(i = 1;i<=tempans;++i){    if(i 属于 集合tempans && i 属于 集合Sj && i是M*α个以上集合的子集)    then ans++；}4.读入用cin的io流处理*/#include <bits/stdc++.h>#include <string>#include <cstring>#include <sstream>#include <algorithm>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<<i);    }    long long ans = 0;    for(int i = 1;i<=tempans;++i){        if((i & tempans)!=i) continue;        long long temp = 0;        for(int j = 0;j<m;++j){            if((i & S[j]) == i)                temp++;            if(temp >= minnum) {                ans++;break;            }        }    }    printf("%lld\n",ans);    return 0;}