文章标题 POJ 2441 ：Arrange the Bulls（状压DP）

Arrange the Bulls

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
题意：有n只牛，m个位置，每只牛有自己喜欢的k个位置，当有一只占据了一个位置，这个位置就不能有其他的牛，现在要我们求出这m个位置分配给n只牛有多少方案数。
分析：显然，当n > m时，方案数为0；由题意可以知道m<=20,所以可以用状态压缩，用dp[i][j]表示前i只牛所形成的状态为j时的方案数，但是直接开开不下，然后我们可以发现类似揹包问题直接从大往小的状态下可以直接开一个一维数组就行了。
代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <queue>
#include <set>
#include <map>
#include <algorithm>
#include <math.h>
#include <vector>
using namespace std;
typedef long long ll;

const int mod=1e9+7;
const int maxn=(1<<20)+10;

int n,m;
int dp[maxn];
int k,tmp;
int mp[25][25];

int main()
{
    while (scanf ("%d%d",&n,&m)!=EOF){
        for (int i=0;i<n;i++){
            scanf ("%d",&k);
            while (k--){
                scanf ("%d",&tmp);
                tmp--;
                mp[i][tmp]=1;
//标记能在tmp这个位置
            }
        }
        if (n>m){
            printf ("0\n");
            continue;
        }
        memset (dp,0,sizeof (dp));
        dp[0]=1;
        for (int i=0;i<n;i++){
            for (int j=(1<<m)-1;j>=0;j--){
                if (dp[j]==0)continue;//表示前i-1只牛没有形成这个状态
                for (int k=0;k<m;k++){//枚举m个场地 
                    if((1<<k)&j)continue;//如果k这个位置已经有人占据了
                    if (mp[i][k]==0)continue;//如果第i只牛不能在第k个位置就跳过
                    int pos=j|(1<<k);
                    dp[pos]+=dp[j]; 
                } 
                dp[j]=0;//表示前i只已经没有这种状态了 
            }
        }
        int ans=0;
        for(int i=0;i<(1<<m);i++){
//求值
            ans+=dp[i];
        } 
        printf ("%d\n",ans);
    }
    return 0;
}