状压DP POJ 3254 Corn Fields

题目传送门

 /*
     状态压缩DP：先处理硬性条件即不能种植的，然后处理左右不相邻的，
     接着就是相邻两行查询所有可行的种数并累加

     写错一个地方差错N久:)
     详细解释：http://www.tuicool.com/articles/JVzMVj
 */
 #include <cstdio>
 #include <iostream>
 #include <algorithm>
 #include <cmath>
 #include <cstring>
 #include <string>
 #include <map>
 #include <set>
 using namespace std;

 const int MAXN = ;
 const int INF = 0x3f3f3f3f;
 const int MOD = 1e8;
 int dp[MAXN][ << MAXN];
 int st[ << MAXN];
 int rem[ << MAXN];

 int solve(int m)
 {
     int cnt = ;
     int tot =  << m;
     for (int i=; i<tot; ++i)
     {
         if ((i & (i << )) == )
         {
             st[++cnt] = i;
         }
     }

     return cnt;
 }

 void work(int n, int m, int cnt)
 {
     memset (dp, , sizeof (dp));
     for (int i=; i<=cnt; ++i)
     {
         if ((st[i] & rem[]) == )
         {
             dp[][i] = ;
         }
     }

     for (int i=; i<=n; ++i)
     {
         for (int k=; k<=cnt; ++k)
         {
             if ((st[k] & rem[i]) == )
             {
                 for (int j=; j<=cnt; ++j)
                 {
                     if (((st[j] & rem[i-]) == ) && ((st[j] & st[k]) == ))
                         dp[i][k] = (dp[i][k] + dp[i-][j]) % MOD;
                 }
             }
         }
     }

     int ans = ;
     for (int i=; i<=cnt; ++i)
     {
         ans = (ans + dp[n][i]) % MOD;
     }

     printf ("%d\n", ans);
 }

 int main(void)        //POJ 3254 Corn Fields
 {
     #ifndef ONLINE_JUDGE
     freopen ("F.in", "r", stdin);
     #endif

     int n, m;
     while (~scanf ("%d%d", &n, &m))
     {
         int cnt = solve (m);

         memset (rem, , sizeof (rem));
         for (int i=; i<=n; ++i)
         {
             int x;
             for (int j=; j<=m; ++j)        //看错n和m
             {
                 scanf ("%d", &x);
                 if (x == )
                 {
                     rem[i] = rem[i] | ( << (m-j));
                 }
             }
         }

         work (n, m, cnt);
     }

     return ;
 }