3006 The Number of set

最新推荐文章于 2022-06-14 10:27:47 发布

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本文探讨了一种利用位运算解决集合组合的问题。给定多个整数集合，每个元素范围限定，通过位运算来高效计算所有可能的新集合总数。文章详细介绍了实现思路与代码示例。

The Number of set

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1199 Accepted Submission(s): 735

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

Sample Output

15
2

第一次接触位运算的题，一个集合内数是多少就将第几位置为1，然后用这个组成的数作为数组下标记录该集合，并置为1，然后遍历合并，

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
const int maxn=1<<14+5;
int a[maxn];
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        memset(a,0,sizeof(a));
        for(int i=1;i<=n;i++)
        {
            int k,ss=0;
            scanf("%d",&k);

            while(k--)
            {
                int s;
                scanf("%d",&s);
               // ss=ss|1<<(s-1);
                ss=ss+pow(2,s-1);
            }
            //cout<<ss<<endl;
            a[ss]=1;
            for(int i=1;i<=1<<m;i++)
            {
                if(a[i])
                    a[i|ss]=1;
            }
         }
         int sum=0;
         for(int i=1;i<=1<<m;i++)
            if(a[i]) sum++;
         printf("%d\n",sum);
     }
    return 0;
}