poj 2441 Arrange the Bulls (状压dp)

在续挑战这本书,看了下状压但是不是很懂,因为老感觉他那几重循环有好多地方是有问题的,但是也不太清楚.
这一题是练习题第一题,然后我想了两个小时才出来,然后发现时间是人家十倍,空间是两倍,就很难过了.
大概我的思维是独一份的把,独一份的蠢..
dp[s][time] 在上一次time(滚动数组)之后,分配方式为s的方法有多少种.

/*  xzppp  */#include <iostream>#include <vector>#include <cstdio>using namespace std;#define FFF freopen("in.txt","r",stdin);freopen("out.txt","w",stdout);#define lson MAXN,m,rt<<1#define rson m+1,r,rt<<1|1#define MP make_pair#define PB push_backtypedef long long  ll;typedef unsigned long long ULL;const int MAXN = 20;const int MAXM = 20;const int INF = 0x7fffffff;const int MOD = 1e9+7;int dp[1<<MAXN][2];vector<int > G[MAXN+17];int main(int argc, char const *argv[]){    //FFF    int n,m;    scanf("%d%d",&n,&m);    for (int i = 0; i < n; ++i)    {        int p;        scanf("%d",&p);        for (int j = 0; j < p; ++j)        {            int temp;            scanf("%d",&temp);            temp--;            G[i].PB(temp);        }    }    int ans = 0;    for (int i = 0; i < G[0].size(); ++i)        dp[1<<G[0][i]][0] = 1;    int now = 0,last = 1;    for (int i = 1; i < n; ++i)    {        swap(now,last);        for (int s = (1<<m)-1; s > -1; --s) dp[s][now] = 0;        for (int j = 0; j < G[i].size(); ++j)            for (int s = (1<<m)-1; s > -1; --s)                if(!((s>>G[i][j])&1)&&dp[s][last]>0)                {                    dp[s|(1<<G[i][j])][now] += dp[s][last];                    if(i==n-1)                        ans+=dp[s][last];                }    }    printf("%d\n",ans);    return 0;}