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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題意：

給你n個集合。集合中均為數字且數字的范圍在[1,m]內。m<=14。現在問用這些集合能組成多少個集合自己本身也算。

思路：

開始有點無頭緒。一看到m范圍就樂了。正好用二進制壓縮。第i為1表示集合裡有i這個數。然後背包就行了。

詳細見代碼：

#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;
}