poj 3254 Corn Fields (狀態壓縮DP)

poj 3254 Corn Fields (狀態壓縮DP)

Corn Fields Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 8041 Accepted: 4287

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.

思路：首先的得到相鄰格子不能同時取的信息，然後，對於不能種的地也要捨去，最後求得不和上一行沖突的狀態數總和。

#include"stdio.h"
#include"string.h"
#include"iostream"
#include"algorithm"
#include"math.h"
#include"vector"
using namespace std;
#define LL __int64
#define N 13
const int mod=100000000;
const int M=1<<12;
int g[N][N];
int num[M],a[N][M];
int dp[N][M];  //存儲每個格子對應的取法總數
int n,m,lim;   //每一行的和就是以該行為土地邊界的答案
void inti()   //初始化數組，相鄰兩個格子不能同時取
{
    int i,k;
    lim=1<<12;
    for(i=k=0;i