计蒜客 2017 ACM-ICPC 亚洲区（南宁赛区）网络赛 M. Frequent Subsets Problem (位压缩)

The frequent subset problem is defined as follows. Suppose $U$={1, 2,$\dots$,N} is the universe, and S1S_{1}S​1​​,S2S_{2}S​2​​,$\dots$,SMS_{M}S​M​​ are $M$ sets over $U$. Given a positive constant $\alpha$,$0<\alpha \le 1$, a subset $B$ ($B\ne 0$) is α-frequent if it is contained in at least $\alpha M$ sets of S1S_{1}S​1​​,S2S_{2}S​2​​,$\dots$,SMS_{M}S​M​​, i.e. ∣{i:B⊆Si}∣≥αM\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$U=\{1,2,3,4,5\}$,$M=3$,$\alpha =0.5$, and S1={1,5}S_{1}=\{1, 5\}S​1​​={1,5},S2={1,2,5}S_{2}=\{1,2,5\}S​2​​={1,2,5},S3={1,3,4}S_{3}=\{1,3,4\}S​3​​={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 $\alpha$, where $N$ is a positive integers, and $\alpha$ 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}S​i​​,$1\le i\le M$ . Your program should be able to handle $N$ up to $20$ and $M$ up to $50$.

Output Format

The number of $\alpha$-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

题意:

有M个全集U下的子集S1....SM,问你存在多少个子集被k个Si包含（即该子集是Si的子集）且k>=(M*a),全集就是{1,2,3,...,n}

解析:

因为n最大只有20个，所以可以用二进制表示。一个n答案（子集）最多有((1<<n)-1)个，就枚举答案，再将每一个答案与每一个集合里的元素（二进制表示）比较，若包含次数>=Ma,说明该答案满足条件

#include<cstdio>#include<cstring>#include<cmath>#include<set>#include<vector>using namespace std;const int N = 30;const int M = 60;int mm[M];int n;int ans;int Ma;int main(){double a;scanf("%d%lf",&n,&a);getchar();int line=0;char str[1000];while(gets(str)!=NULL){line++;int temp=0;for(int i=0;str[i]!='\0';i++){if(str[i]==' '||str[i]=='\0'){mm[line]=mm[line]|(1<<(temp-1));temp=0;}else{temp=temp*10+str[i]-'0';}}if(temp) {mm[line]=mm[line]|(1<<(temp-1));}//if(line==5) break;}Ma=(int)((a*line)+0.999999999);ans=0;for(int i=1;i<(1<<n);i++){int fi=0;for(int j=1;j<=line;j++){if((i&mm[j])==i) fi++;if(fi>=Ma) break;}if(fi>=Ma) ans++;}    printf("%d\n",ans);    return 0;}