POJ 2441 Arrange the Bulls 状压dp

题目链接：

Arrange the Bulls

Time Limit: 4000MS
Memory Limit: 65536K #### 问题描述 > 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.

输入

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.

输出

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

样例输入

3 4
2 1 4
2 1 3
2 2 4

样例输出

4

题意

有n只牛，m个体育场，每只牛只会去若干个自己喜欢的体育场，现在要分配n只牛到n个不同的体育场，问总共有多少种分配方案。

题解

dp[i][j]表示分配前i只牛在状态为j的体育场的总方案数。

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf

typedef int LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=10000000000000000LL;
const double eps=1e-9;

const double PI = acos(-1.0);

//start----------------------------------------------------------------------

const int maxn=20;

int dp[2][1<<maxn];
///sumv统计二进制中1的个数
int sumv[1<<maxn];
int n,m;

bool mp[maxn][maxn];

void pre(){
    clr(sumv,0);
    rep(i,0,(1<<maxn)){
        rep(j,0,maxn){
            if(i&(1<<j)) sumv[i]++;
        }
    }
}

void init(){
    clr(mp,0);
}

int main() {
    pre();
    while(scf("%d%d",&n,&m)==2&&n) {
        init();
        rep(i,0,n){
            int cnt; scf("%d",&cnt);
            while(cnt--){
                int v; scf("%d",&v);
                v--;
                mp[i][v]=true;
            }
        }

        int pre=0,cur=1;
        clr(dp[cur],0);

        ///初始化
        for(int j=0;j<m;j++){
            if(mp[0][j]){
                dp[cur][1<<j]=1;
            }
        }

        for(int i=1;i<n;i++){
            swap(pre,cur);
            clr(dp[cur],0);
            for(int j=0;j<m;j++){
                if(mp[i][j]==0) continue;
                for(int k=0;k<(1<<m);k++){
                    if(k&(1<<j)) continue;
                    dp[cur][k^(1<<j)]+=dp[pre][k];
                }
            }
        }

        int ans=0;
        for(int i=0;i<(1<<m);i++){
            if(sumv[i]==n) ans+=dp[cur][i];
        }

        prf("%d
",ans);

    }
    return 0;
}

//end-----------------------------------------------------------------------