poj 2441 Arrange the Bulls（状压DP入门）

Arrange the Bulls

Time Limit: 4000MS		Memory Limit: 65536K
Total Submissions: 3509		Accepted: 1344

Description

Farmer Johnson's Bulls love playing basketball very much. But none of them would like to play basketball with the other bulls because they believe that the others are all very weak. Farmer Johnson has N cows (we number the cows from 1 to N) and M barns (we
number the barns from 1 to M), which is his bulls' basketball fields. However, his bulls are all very captious, they only like to play in some specific barns, and don’t want to share a barn with the others.

So it is difficult for Farmer Johnson to arrange his bulls, he wants you to help him. Of course, find one solution is easy, but your task is to find how many solutions there are.

You should know that a solution is a situation that every bull can play basketball in a barn he likes and no two bulls share a barn.

To make the problem a little easy, it is assumed that the number of solutions will not exceed 10000000.
Input

In the first line of input contains two integers N and M (1 <= N <= 20, 1 <= M <= 20). Then come N lines. The i-th line first contains an integer P (1 <= P <= M) referring to the number of barns cow i likes to play in. Then follow P integers, which give the
number of there P barns.
Output

Print a single integer in a line, which is the number of solutions.
Sample Input

3 4
2 1 4
2 1 3
2 2 4

Sample Output

4

刚开始学习状压DP，有理解不对的还忘指正，谢谢。
参考的代码：/article/1753169.html
参考资料：http://www.hankcs.com/program/algorithm/poj-2441-arrange-the-bulls.html
我对代码的理解

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int map[25][25];
int dp[(1<<20)+100];
int main()
{
int n,m;
while(~scanf("%d%d",&n,&m))
{
int p;
memset(map,0,sizeof(map));
for(int i=1;i<=n;i++)
{
scanf("%d",&p);
for(int j=1;j<=p;j++)
{
int x;
scanf("%d",&x);
x--;//1移x-1位就是第x个1
map[i][x]=1;
}
}
if(n>m)
{
printf("0\n");
continue;
}
memset(dp,0,sizeof(dp));
dp[0]=1;//全部都不选
for(int i=1;i<=n;i++)
{
for(int j=(1<<m)-1;j>=0;j--)//枚举子集
{
if(!dp[j])//如果该子集不存在,跳过
continue;
for(int k=0;k<m;k++)//向子集中添加元素
{
if(((1<<k)&j)!=0)//该元素已经在子集中了就跳过
continue;
if(map[i][k]==0)//不能取该元素
continue;
int temp=((1<<k)|j);//添加元素后的新的值
dp[temp]+=dp[j];//添加元素
}
dp[j]=0;//这个子集就不存在了。
}
}
long long ans=0;
for(int i=0;i<(1<<m);i++)
ans+=dp[i];
printf("%I64d\n",ans);
}
return 0;
}