poj3254：Corn Field

Corn Fields

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 10149		Accepted: 5366

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 3
1 1 1
0 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.

 

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=>1.首先是排除格子相邻的情况，if(x&x<<1)  return 0;return 1;

=>2.接着是读入，将状态保存下来，如果x==0=>cur[i]+=(1<<n-j);

=>3.然后是处理边界，f(x&cur(1))?return 0;return 1;

=>4.接着就是动规了，如果这种状态没问题，枚举，如果下一行的也没问题，如果这两行没问题，累加；

=>5.每一行的一个状态等于上一行的所有和该状态切合的状态的ans的和；end：将最后一行的累加即可；

=>6.打得慢的原因有一个就是不知道要<<多少位，if移i位，实际移i-1位，同时逆序的话，就n-i位即可；

=>7.orrrrrrrrrrrrrrrrrz!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!;自己打了一遍结果漏加了两个括号bug了很久，切记切记。括号尽管加，不要省！！！

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ps：第一次做状压dp几乎是照着代码打的，不过新手嘛大概可以原谅（。。。）；