The frequent subset problem is defined as follows. Suppose UU={1, 2,\ldots…,N} is the universe, and S_{1}S 1 , S_{2}S 2 ,\ldots…,S_{M}S M are MM sets over UU. Given a positive constant \alphaα, 0<\alpha \leq 10<α≤1, a subset BB (B \neq 0B≠0) is α-frequent if it is contained in at least \alpha MαM sets of S_{1}S 1 , S_{2}S 2 ,\ldots…,S_{M}S M , i.e. \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}U={1,2,3,4,5}, M=3M=3, \alpha =0.5α=0.5, and S_{1}={1, 5}S 1 ={1,5}, S_{2}={1,2,5}S 2 ={1,2,5}, S_{3}={1,3,4}S 3 ={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}S i , 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
太神奇了，对DP的了解为0，网上说这是状压DP，我认为这个位运算操作表示一个集合太牛逼了，暴力，先把所有集合的用一个数来表示（110，为5，表示这个集合中有2，3），判断一个集合是否是另一个集合的子集，i&a[j])==i，遍历所有子集，找出有几个集合是大于a*m个集合的子集。。 是该看看DP了
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> using namespace std; int a[55]; int main() { memset(a,0,sizeof a); int m=0,n; double p; scanf("%d%lf",&n,&p); int num; char ch; while(scanf("%d%c",&num,&ch)!=EOF) { a[m]+=(1<<(num-1)); if(ch=='\n')m++; } int ans=0; int sta=ceil(m*p); for(int i=1;i<(1<<n);i++) { int sum=0; for(int j=0;j<m;j++) { if((i&a[j])==i)sum++; } if(sum>=sta)ans++; } cout<<ans<<endl; return 0; }