hdu 3006 The Number of set

最新推荐文章于 2019-11-14 20:09:17 发布

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本文深入探讨了编程领域中常见的复杂问题解决策略，涵盖了从算法优化到数据结构选择，再到软件架构设计等多个方面。通过实例分析，旨在为开发者提供一套全面的技术解决方案，帮助他们更高效地应对编程挑战。

The Number of set

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 832 Accepted Submission(s): 511

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

Source

2009 Multi-University Training Contest 11 - Host by HRBEU

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位运算....

#include <stdio.h>
#include <string.h>
int s[1<<15];
int main()
{
    int m,n,t,data;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        memset(s,0,sizeof(s));
        while(n--)
        {
            int ans=0;
            scanf("%d",&t);
            for(int i=1;i<=t;i++)
            {
                scanf("%d",&data);
                ans=ans|(1<<(data-1));
            }
            s[ans]=1;
            for(int i=0;i<=(1<<14);i++)
            {
                if(s[i])
                s[ans|i]=1;
            }
        }
        int num=0;
        for(int i=0;i<(1<<14);i++)
        if(s[i])
        num++;
        printf("%d\n",num);
    }
    return 0;
}