Frequent Subsets Problem

The frequent subset problem is defined as follows. Suppose UU={1, 2,ldots…,N} is the universe, and S_{1}S1​, S_{2}S2​,ldots…,S_{M}SM​are MM sets over UU. Given a positive constant alphaα, 0<alpha leq 10<α≤1, a subset BB (B eq 0B≠0) is α-frequent if it is contained in at least alpha MαM sets of S_{1}S1​, S_{2}S2​,ldots…,S_{M}SM​, i.e. left | left { i:Bsubseteq S_{i} ight } ight | geq alpha M∣{i:B⊆Si​}∣≥αM. The frequent subset problem is to find all the subsets that are α-frequent. For example, let U={1, 2,3,4,5}U={1,2,3,4,5}, M=3M=3, alpha =0.5α=0.5, and S_{1}={1, 5}S1​={1,5}, S_{2}={1,2,5}S2​={1,2,5}, S_{3}={1,3,4}S3​={1,3,4}. Then there are 33 α-frequent subsets of UU, which are {1}{1},{5}{5} and {1,5}{1,5}.

Input Format

The first line contains two numbers NN and alphaα, where NN is a positive integers, and alphaα 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 + 1i+1 contains S_{i}Si​, 1 le i le M1≤i≤M . Your program should be able to handle NN up to 2020 and MM up to 5050.

Output Format

The number of alphaα-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

这几天天天写dp，但是基本上是碰见一天写一题先是状态压缩的dp，写了两道有换成了区间dp的题，后来又换成了概率的和数位dp，现在写一些简单的状态压缩dp已经没有什么问题了，决定现在先学会数位dp然后开始努力刷dp题（当然是各种dp），希望能在一个整体上提高。。

#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#include <queue>
#include <vector>
#include <math.h>
#include <string.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 20
#define mid 1e-8
#define LL long long
/********************************************************/
int Map[55];
int main()
{
    int T, t=1, m, n, x;
    double a;
    char ch;

    scanf("%d%lf", &n, &a);
    memset(Map, 0, sizeof(Map));

    m=0;
    while(scanf("%d", &x)!=EOF)
    {
        ch=getchar();
        Map[m]=Map[m]|(1<<(x-1));
        if(ch=='
') m++;
    }

    a=ceil(a*m);
    x=0;
    for(int i=1;i<(1<<n);i++)
    {
        int sum=0;
        for(int j=0;j<m;j++)
        {
            if((Map[j]&i)==i)
            {
                sum++;
            }
        }
        if(sum>=a) x++;
    }

    printf("%d
", x);
    return 0;
}