计蒜客 Frequent Subsets Problem(状态压缩)
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The frequent subset problem is defined as follows. Suppose U={1, 2,…,N} is the universe, and S1, S2,…,SMare M sets over U. Given a positive constant α, 0<α≤1, a subset B (B≠0) is α-frequent if it is contained in at least αM sets of S1, S2,…,SM, i.e. ∣{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}, M=3, α=0.5, and S1={1,5}, S2={1,2,5}, S3={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 α, where N is a positive integers, and 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 Si, 1≤i≤M . Your program should be able to handle N up to 20 and M up to 50.
Output Format
The number of α-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
题目来源
2017 ACM-ICPC 亚洲区(南宁赛区)网络赛
#include<iostream>#include<stdio.h>#include<string.h>using namespace std;int s[55];int main(){memset(s,0,sizeof(s));int n,m=0;double a;scanf("%d%lf",&n,&a);int x;char p;while(scanf("%d%c",&x,&p)!=EOF){s[m]+=(1<<(x-1));if(p=='\n')m++;}int g=m*a+(bool)(m*a-(int)(m*a)); //代表至少为g个给定集合的子集 int ans=0;for(int i=1;i<(1<<n);i++) //二进制表示状态,枚举所有集合 {int sum=0;for(int j=0;j<m;j++) //跑遍所有给定m个集合判断是不是子集关系 {if((i&s[j])==i)sum++;}if(sum>=g)ans++;}printf("%d\n",ans);}
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