题目链接

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.

#include <iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<map>
#include<queue>
#include<set>
#include<cmath>
#include<stack>
#include<string>
const int maxn=1e4+10;
const int mod=100000000;
const int inf=1e8;
#define me(a,b) memset(a,b,sizeof(a))
#define lowbit(x) x&(-x)
typedef long long ll;
using namespace std;
int dp[20][1<<12],state[maxn],fit[15],tot;
int n,m;
bool check(int x)
{
    if(x&(x<<1))// x左移1后每个位置都和原先后边的位置进行&运算， 若有相邻的1， 则会得到1；
        return 0;
    return 1;
}
bool check1(int x,int i)
{
    if(x&fit[i])//fild存的是土地状态取反；若x&fild[i]==1表示在两数的某位置都为1；
        return 0;
    return 1;
}
void init()//提前存下可行状态
{
    tot=0;
    for(int i=0; i<(1<<n); i++)
        if(check(i))
            state[++tot]=i;
}
int main()
{
    scanf("%d%d",&m,&n);
    init();
    for(int i=1; i<=m; i++)
    {
        fit[i]=0;
        for(int j=1; j<=n; j++)
        {
            int x;scanf("%d",&x);
            if(x==0)
                fit[i]+=(1<<(n-j));//反着存，0表示可以放，1表示不能放
        }
    }
    for(int i=1; i<=tot; i++)//判断第一行；找出第一行可行方案；
        if(check1(state[i],1))
            dp[1][i]=1;
    for(int i=2;i<=m;i++)
    {
        for(int j=1;j<=tot;j++)
        {
            if(!check1(state[j],i))//该位置与当前列冲突。
                continue ;
            for(int k=1;k<=tot;k++)
            {
                if(!check1(state[k],i-1))//当前行与上一行冲突
                    continue ;
                if(state[j]&state[k])//同一行两个位置冲突。
                    continue ;
                dp[i][j]=(dp[i][j]+dp[i-1][k])%mod;
            }
        }
    }
    int ans=0;
    for(int i=1;i<=tot;i++)
        ans=(ans+dp[m][i])%mod;
    cout<<ans<<endl;
    return 0;
}