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.
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
Recommend
题意：
给你n个集合。集合中均为数字且数字的范围在[1,m]内。m<=14。现在问用这些集合能组成多少个集合自己本身也算。
思路：
开始有点无头绪。一看到m范围就乐了。正好用二进制压缩。第i为1表示集合里有i这个数。然后背包就行了。
详细见代码：
#include<stdio.h>
#include<iostream>
#include<string.h>
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<n;i++)
{
s[i]=0;
scanf("%d",&k);
for(j=0;j<k;j++)
{
scanf("%d",&tp);
s[i]=s[i]|base[tp-1];
}
}
for(i=0;i<n;i++)
for(j=base[m]-1;j>=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;
}