hdu 3006 The Number of set(思维+壮压DP)
2014-05-08 11:24:05         来源：hdu 3006 The Number of set(思维+壮压DP)

# The Number of set

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1056 Accepted Submission(s): 655

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 4
1 1
1 2
1 3
1 4
2 4
3 1 2 3
4 1 2 3 4```

Sample Output
```15
2```

Source 2009 Multi-University Training Contest 11 - Host by HRBEU
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```#include
#include
#include
using namespace std;

const int maxn=100010;
//typedef __int64 ll;
int dp[1<<15],s[110],base[15];
int main()
{
int n,m,i,j,k,tp,ans;

base[0]=1;
for(i=1;i<=15;i++)
base[i]=base[i-1]<<1;
while(~scanf("%d%d",&n,&m))
{
memset(dp,0,sizeof dp);
dp[0]=1,ans=0;
for(i=0;i=0;j--)
if(dp[j])
dp[j|s[i]]=1;
for(i=base[m]-1;i>=1;i--)
if(dp[i])
ans++;
printf("%d\n",ans);
}
return 0;
}
```