HDUOJ3006 The Number fo set

 

The Number of set

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 413    Accepted Submission(s): 244

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

 

 

 

【解题思路】这是一道经典的状态压缩题，把一个集合看做一个int用bit位来代表集合的元素，之后编程也有点小技巧不然会超时。

 

【源码】

#include<iostream>using namespace std;#define max (1<<14) + 5bool a[max];int main(){ //freopen("test.txt","r",stdin); int n,m,ans,num,i,j,tmp,cl; while(cin>>n>>m) { memset(a,0,sizeof(a)); for(i=0;i<n;i++) { cin>>num; for(j=0,cl=0;j<num;j++) { cin>>tmp; cl = cl|(1<<tmp-1); } a[cl] = 1; for(j=0;j<max;j++) if(a[j]) a[cl|j] = 1; } for(i=0,ans =0;i<max;i++) if(a[i])ans++; cout<<ans<<endl; } return 0;}