Corn Fields

Time Limit: 2000MS	Memory Limit: 65536K
Total Submissions: 14837	Accepted: 7768

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

【題目大意】一個矩陣裡有很多格子，每個格子有兩種狀態，可以放牧和不可以放牧，可以放牧用1表示，否則用0表示，在這塊牧場放牛，要求兩個相鄰的方格不能同時放牛，即牛與牛不能相鄰。問有多少种放牛方案(一頭牛都不放也是一種方案)

分析：狀態壓縮DP，時間複雜度O（n* 2^m *2^m） , 用 x & （x<<1）來判斷相鄰位置是否有都為1的情況，

【狀態表示】dp[state][i]：在狀態為state時，到第i行符合條件的可以放牛的方案數

【狀態轉移方程】dp[state][i] =Sigma dp[state'][i-1] （state'為符合條件的所有狀態）

【DP邊界條件】首行放牛的方案數dp[state][1] =1(state符合條件) OR 0 (state不符合條件)

程式碼：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <queue>
#define mem(p,k) memset(p,k,sizeof(p));
#define rep(a,b,c) for(int a=b;a<c;a++)
#define pb push_back
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define inf 0x6fffffff
#define ll long long
using namespace std;
const int N=30010;
const int mod=100000000;
int n,m,num,tol;
int mapp[15],plan[10000];
int dp[15][10000];
int main(){

    while(~scanf("%d%d",&n,&m)){
        tol=0;
        for(int i=0;i<(1<<m);i++){
            if(i&(i<<1))continue;
            plan[tol++]=i;
        }
        for(int i=1;i<=n;i++){
            mapp[i]=0;
            for(int j=0;j<m;j++){
                scanf("%d",&num);
                if(!num)mapp[i]+=(1<<j);
            }
        }
        for(int i=0;i<tol;i++){
            if(plan[i] & mapp[1])dp[1][i]=0;
            else dp[1][i]=1;
        }
        for(int i=2;i<=n;i++){
            for(int j=0;j<tol;j++){
                if(plan[j] & mapp[i])continue;
                for(int k=0;k<tol;k++){
                    if(plan[j] & plan[k] || plan[k] & mapp[i-1])continue;
                    dp[i][j]+=dp[i-1][k];
                    dp[i][j]%=mod;
                }
            }
        }
        int ans=0;
        for(int i=0;i<tol;i++)ans=(ans+dp[n][i])%mod;
        printf("%d\n",ans);
    }

    return 0;
}

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