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

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The frequent subset problem is defined as follows. Suppose $U$ ={1, 2, $\ldots$ ,N} is the universe, and $S_{1}$ , $S_{2}$ , $\ldots$ , $S_{M}$ are $M$ sets over $U$ . Given a positive constant $\alpha$ , $0<\alpha \leq 1$ , a subset $B$ ( $\neq 0$ ) is α-frequent if it is contained in at least $\alpha M$ sets of $S_{1}$ , $S_{2}$ , $\ldots$ , $S_{M}$ , i.e. $\left | \left \{ i:B\subseteq S_{i} \right \} \right | \geq \alpha M$ . 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 $S_{1}=\{1, 5\}$ , $S_{2}=\{1,2,5\}$ , $S_{3}=\{1,3,4\}$ . 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 $S_{i}$ , $\le i \le M$ . 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

题目来源

2017 ACM-ICPC 亚洲区（南宁赛区）网络赛

题意：给你一个N和a，所有U的子集里（U={1,2,3,4,...N}），有多少个是频繁子集。频繁子集就是给定M个集合，该子集被这些集合包含的次数要大于 a*M。

解题思路：一开始没注意N最大20……瞎几把搞……一直搞不出来……后来发现之后……直接暴力状压就可以了………我甚至怀疑直接用stl里的set都能过。

#include<iostream>#include<algorithm>#include<math.h>using namespace std;typedef long long int ll;int S[100];//给你的集合int main(){    int m=0;    int n;    double a;    scanf("%d%lf",&n,&a);    int num;    char ch;    while(~scanf("%d%c",&num,&ch))    {        S[m]+=(1<<(num-1));        if(ch=='\n')            m++;    }    int ans=0;    int nnum=ceil(m*a);    //枚举所有子集    for(int i=1;i<(1<<n);i++)    {        double sum=0;        for(int j=0;j<m;j++)        {            if((i&S[j])==i)                sum++;        }        if(sum>=nnum)            ans++;    }    printf("%d\n",ans);    return 0;}

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