POJ 3254 Corn Fields（状压DP）

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.

题意：农夫约翰有一块n*m的草地，1表示可以放牛，0表示不能放牛，约翰不会同时在相邻的草地上放牛，问共有多少种放牛的方案，答案对1e8取模。

题解：枚举所有合法状态

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int a[15];
int dp[15][1<<15];
const int mod=1e8;
bool check(int x)
{
    if(x&(x>>1))//有两个1相邻，结果不为0
        return 0;

    return 1;//没有两个1相邻，结果为0
}
bool check2(int x,int y)
{
    if(x&y) //与上一行冲突，结果不为0
        return 0;

    return 1;//与上一行不冲突，结果为0
}
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        memset(a,0,sizeof(a));
        memset(dp,0,sizeof(dp));
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                int x;
                scanf("%d",&x);
                if(x==1)
                {
                    a[i]|=(1<<j);//每一行的状态
                }
            }
        }
        for(int i=0;i<(1<<m);i++)
        {
            if((a[0]|i)==a[0]&&check(i))
            {
                dp[0][i]=1;//第一行的可行方案
            }
        }
        for(int i=1;i<n;i++)
        {
            for(int j=0;j<(1<<m);j++)
            {
                if((a[i]|j)==a[i]&&check(j))//当前状态满足
                {
                    for(int k=0;k<(1<<m);k++)
                    {
                        if(check2(j,k))//与上一行不冲突
                        {
                            dp[i][j]=(dp[i][j]+dp[i-1][k])%mod;//加上上一行的方案数
                        }
                    }
                }
            }
        }
        int ans=0;
        for(int i=0;i<(1<<m);i++)
        {
            ans=(ans+dp[n-1][i])%mod;
        }
        printf("%d\n",ans);
    }
}