Frequent Subsets Problem
The frequent subset problem is defined as follows. Suppose U={1, 2,…\ldots…,N} is the universe, and S1S_{1}S1, S2S_{2}S2,…\ldots…,SMS_{M}SM are MMM sets over UUU. Given a positive constant α\alphaα, 0<α≤10<\alpha \leq 10<α≤1, a subset BBB (B≠0B \neq 0B≠0) is α-frequent if it is contained in at least αM\alpha MαM sets of S1S_{1}S1, S2S_{2}S2,…\ldots…,SMS_{M}SM, i.e. ∣{i:B⊆Si}∣≥αM\left | \left \{ i:B\subseteq S_{i} \right \} \right | \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\}U={1,2,3,4,5}, M=3M=3M=3, α=0.5\alpha =0.5α=0.5, and S1={1,5}S_{1}=\{1, 5\}S1={1,5}, S2={1,2,5}S_{2}=\{1,2,5\}S2={1,2,5}, S3={1,3,4}S_{3}=\{1,3,4\}S3={1,3,4}. Then there are 333 α-frequent subsets of UUU, which are {1}\{1\}{1},{5}\{5\}{5} and {1,5}\{1,5\}{1,5}.
Input Format
The first line contains two numbers N and α\alpha α, where N is a positive integers, and α\alpha α 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}Si, 1≤i≤M1 \le i \le M1≤i≤M . Your program should be able to handle NNN up to 202020 and MMM up to 505050.
Output Format
The number of α\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
题目来源
求在m个集合中出现次数大于a*m的子集数量。n不大与20 最大1<<20 不会超时，直接暴力可以做。
1 #include <bits/stdc++.h> 2 #define ll long long 3 using namespace std; 4 const int N = 110; 5 int a[N]; 6 int main() { 7 int n, num = 0, m = 0; 8 double aa; 9 char c; 10 scanf("%d %lf", &n, &aa); 11 getchar(); 12 while((c = getchar()) != EOF) { 13 int ans = c - '0'; 14 while((c = getchar()) != '\n') { 15 if(c >= '0' && c <= '9') { 16 ans = ans * 10 + c - '0'; 17 }else { 18 a[m] += 1 << (ans - 1); 19 ans = 0; 20 } 21 } 22 a[m] += 1 << (ans - 1); 23 m ++; 24 } 25 int ma = ceil(aa*m); 26 int cnt = 0; 27 for(int i = 1; i < (1<<n); i ++) { 28 int num = 0; 29 for(int j = 0; j < m; j ++) { 30 if((i & a[j]) == i) num ++; 31 } 32 if(num >= ma) cnt++; 33 } 34 printf("%d\n",cnt); 35 return 0; 36 }
