题意：
给n个集合，集合中的数不会大于m(1<=n<=100,1<=m<=14)，求n个集合任意组合，能得到的集合数。
思路：
因为m不大，所以以m为突破口，枚举1到m的所有集合，查询所有集合看能否组成当前枚举出的集合。每个集合通过一个int的二进制去记录其包含的元素，1表示存在，0表示不存在，加速求并集的操作。
代码：
#include <iostream>
#include <iomanip>
#include <algorithm>
#include <cstring>
#include <cctype>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <map>
#include <list>
#include <set>
#include <stack>
#include <queue>
#include <string>
#include <sstream>
#define pb push_back
#define X first
#define Y second
#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())
#define pii pair<int,int>
#define qclear(a) while(!a.empty())a.pop();
#define lowbit(x) (x&-x)
#define sd(n) scanf("%d",&n)
#define sdd(n,m) scanf("%d%d",&n,&m)
#define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)
#define mst(a,b) memset(a,b,sizeof(a))
#define cout3(x,y,z) cout<<x<<" "<<y<<" "<<z<<endl
#define cout2(x,y) cout<<x<<" "<<y<<endl
#define cout1(x) cout<<x<<endl
#define IOS std::ios::sync_with_stdio(false)
#define SRAND srand((unsigned int)(time(0)))
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int uint;
using namespace std;
const double PI=acos(-1.0);
const int INF=0x3f3f3f3f;
const ll mod=998244353;
const double eps=1e-5;
const int maxn=105;
const int maxm=10005;
int n,m;
int s[maxn];
bool vis[20];
void solve()
{
while(~sdd(n,m)){
int ans=0;
mst(s,0);
for(int i=0;i<n;i++){
mst(vis,0);
int nown;
sd(nown);
for(int j=0;j<nown;j++){
int now;
sd(now);
now=1<<(now-1);
s[i]=s[i]|now;
}
}
int lim=1<<m;
for(int i=1;i<lim;i++){
int now1=i;
int now0=i^(lim-1);
int temp=0;
for(int j=0;j<n;j++){
if(s[j]&now0)continue;
else temp=temp|s[j];
}
if(temp==now1)ans++;
}
printf("%d\n",ans);
}
return ;
}
int main()
{
#ifdef LOCAL
freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout);
#else
// freopen("","r",stdin);
// freopen("","w",stdout);
#endif
solve();
return 0;
}