HDU3006-The Number of se

The Number of set

                                                                            Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
                                                                                                       Total Submission(s): 1353    Accepted Submission(s): 835

Problem Description

Given you n sets.All positive integers in sets are not less than 1 and not greater than m.If use these sets to combinate the new set,how many different new set you can get.The given sets can not be broken.

 

Input

There are several cases.For each case,the first line contains two positive integer n and m(1<=n<=100,1<=m<=14).Then the following n lines describe the n sets.These lines each contains k+1 positive integer,the first which is k,then k integers are given. The input is end by EOF.

 

Output

For each case,the output contain only one integer,the number of the different sets you get.

 

Sample Input

4 41 11 21 31 42 43 1 2 34 1 2 3 4

 

Sample Output

152

题目大意：给定n个集合，每个集合都是由大于等于1小于等于m的数字组成，m最大为14。由给出的集合可以组成多少个不同的集合。
输入描述：第一行为n，m，接下来n行，第一个为k，表示该集合的元素个数，接下来k个数表示集合元素。

#include <iostream>#include <algorithm>#include <stdio.h>#include <queue>#include <string.h>using namespace std;int a[1<<16];int main(){    int n,m;    while(~scanf("%d %d",&n,&m))    {        memset(a,0,sizeof a);        for(int i=0;i<n;i++)        {            int k,p,x=0;            scanf("%d",&k);            for(int j=0;j<k;j++)            {                scanf("%d",&p);                x+=(1<<(p-1));            }            a[x]=1;            for(int j=1;j<=(1<<m);j++)            {                if(a[j]) a[j|x]=1;            }        }        int ans=0;        for(int i=1;i<=(1<<m);i++)            if(a[i]) ans++;        printf("%d\n",ans);    }    return 0;}