状压DP POJ 3254 Corn Fields

 

题目传送门

 1 /*  2  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，  3  接着就是相邻两行查询所有可行的种数并累加  4  5  写错一个地方差错N久:)  6  详细解释：http://www.tuicool.com/articles/JVzMVj  7 */  8 #include <cstdio>  9 #include <iostream>  10 #include <algorithm>  11 #include <cmath>  12 #include <cstring>  13 #include <string>  14 #include <map>  15 #include <set>  16 using namespace std;  17  18 const int MAXN = 15;  19 const int INF = 0x3f3f3f3f;  20 const int MOD = 1e8;  21 int dp[MAXN][1 << MAXN];  22 int st[1 << MAXN];  23 int rem[1 << MAXN];  24  25 int solve(int m)  26 {  27 int cnt = 0;  28 int tot = 1 << m;  29 for (int i=0; i<tot; ++i)  30  {  31 if ((i & (i << 1)) == 0)  32  {  33 st[++cnt] = i;  34  }  35  }  36  37 return cnt;  38 }  39  40 void work(int n, int m, int cnt)  41 {  42 memset (dp, 0, sizeof (dp));  43 for (int i=1; i<=cnt; ++i)  44  {  45 if ((st[i] & rem[1]) == 0)  46  {  47 dp[1][i] = 1;  48  }  49  }  50  51 for (int i=2; i<=n; ++i)  52  {  53 for (int k=1; k<=cnt; ++k)  54  {  55 if ((st[k] & rem[i]) == 0)  56  {  57 for (int j=1; j<=cnt; ++j)  58  {  59 if (((st[j] & rem[i-1]) == 0) && ((st[j] & st[k]) == 0))  60 dp[i][k] = (dp[i][k] + dp[i-1][j]) % MOD;  61  }  62  }  63  }  64  }  65  66 int ans = 0;  67 for (int i=1; i<=cnt; ++i)  68  {  69 ans = (ans + dp[n][i]) % MOD;  70  }  71  72 printf ("%d\n", ans);  73 }  74  75 int main(void) //POJ 3254 Corn Fields  76 {  77  #ifndef ONLINE_JUDGE  78 freopen ("F.in", "r", stdin);  79 #endif  80  81 int n, m;  82 while (~scanf ("%d%d", &n, &m))  83  {  84 int cnt = solve (m);  85  86 memset (rem, 0, sizeof (rem));  87 for (int i=1; i<=n; ++i)  88  {  89 int x;  90 for (int j=1; j<=m; ++j) //看错n和m  91  {  92 scanf ("%d", &x);  93 if (x == 0)  94  {  95 rem[i] = rem[i] | (1 << (m-j));  96  }  97  }  98  }  99 100  work (n, m, cnt); 101  } 102 103 return 0; 104 }