poj3254 Corn Fields

Description

Farmer John has purchased a lush new rectangular pasture composed of M by N (1 ≤ M ≤ 12; 1 ≤ N ≤ 12) square parcels. He wants to grow some yummy corn for the cows on a number of squares. Regrettably, some of the squares are infertile and can't be planted. Canny FJ knows that the cows dislike eating close to each other, so when choosing which squares to plant, he avoids choosing squares that are adjacent; no two chosen squares share an edge. He has not yet made the final choice as to which squares to plant.

Being a very open-minded man, Farmer John wants to consider all possible options for how to choose the squares for planting. He is so open-minded that he considers choosing no squares as a valid option! Please help Farmer John determine the number of ways he can choose the squares to plant.

Input

Line 1: Two space-separated integers: M and N 
Lines 2..M+1: Line i+1 describes row i of the pasture with N space-separated integers indicating whether a square is fertile (1 for fertile, 0 for infertile)

Output

Line 1: One integer: the number of ways that FJ can choose the squares modulo 100,000,000.

Sample Input

2 31 1 10 1 0

Sample Output

9

Hint

Number the squares as follows:

1 2 3  4  

There are four ways to plant only on one squares (1, 2, 3, or 4), three ways to plant on two squares (13, 14, or 34), 1 way to plant on three squares (134), and one way to plant on no squares. 4+3+1+1=9.

Source

状态压缩DP。

把每一行的状态用二进制的数表示，0代表不在这块放牛，1表示在这一块放牛。首先很容易看到，每一行的状态要符合牧场的硬件条件，即牛必须放在能放牧的方格上。这样就能排除一些状态。另外，牛与牛之间不能相邻，这样就要求每一行中不能存在两个相邻的1，这样也能排除很多状态。然后就是根据上一行的状态转移到当前行的状态的问题了。必须符合不能有两个1在同一列（两只牛也不能竖着相邻）的条件。这样也能去掉一些状态。然后，上一行的所有符合条件的状态的总的方案数就是当前行该状态的方案数。

于是DP：

【状态表示】dp[state][i]：在状态为state时，到第i行符合条件的可以放牛的方案数

【状态转移方程】dp[state][i] =Sigma dp[state'][i-1] （state'为符合条件的所有状态）

【DP边界条件】首行放牛的方案数dp[state][1] =1(state符合条件) OR 0 (state不符合条件)

(注意一下poj不能用#Include<bits/stdc++.h>头文件)

代码：

<pre name="code" class="cpp">#include <cstdio>#include <cstring>using namespace std;#define mod 100000000int M,N,top = 0;int state[600],num[110];int dp[20][600];int cur[20];inline bool ok(int x){   if(x & x << 1)return 0;   return 1;}void init(){   top = 0;   int total = 1 << N;   for(int i = 0; i < total; i ++)       if(ok(i))state[++ top] = i;   }inline bool fit(int x,int k){   if(x & cur[k])return 0;   return 1;}inline int jcount(int x){   int cnt = 0;   while(x)   {       cnt ++;       x &= (x - 1);   }   return cnt;} int main(){    while(scanf("%d%d",&M,&N) != EOF)    {       init();       memset(dp,0,sizeof(dp));       for(int i = 1; i <= M; i ++){           cur[i] = 0;           int num;           for(int j = 1; j <= N; j ++) {                scanf("%d",&num);               if(num == 0)cur[i] += (1 << (N - j));           }       }       for(int i = 1;i <= top;i++){           if(fit(state[i],1)){                dp[1][i] = 1;           }       }       for(int i = 2; i <= M; i ++)       {           for(int k = 1; k <= top; ++k)<span style="white-space:pre"></span>   {                if(!fit(state[k],i))continue;                for(int j = 1; j <= top ;j ++)<span style="white-space:pre"></span>{                   if(!fit(state[j],i-1))continue;                   if(state[k]&state[j])continue;                    dp[i][k] = (dp[i][k] + dp[i-1][j]) % mod;                }           }       }       int ans = 0;       for(int i = 1; i <= top; i ++)           ans = (ans + dp[M][i])%mod;       printf("%d\n",ans);   }}