2017 ACM-ICPC 亚洲区(南宁赛区)网络赛M题

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;  }