poj 3254 Corn Fields

Corn Fields

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 11000		Accepted: 5773

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：一个n*m的矩阵，每个格子是0或者1，1表示土壤肥沃可以种植草地，0则不可以。在种草地的格子可以放牛，但边相邻的两个格子不允许同时放牛，问总共有多少种放牛的方法？（不放牛也算一种情况）

思路：这是一道状态压缩dp；设dp[i][j]为前i行，第i行状态为j的方法总数，则状态转移方程为：dp[i][j]+=dp[i-1][k](其中j与k不冲突)；

代码如下：

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int MOD = 100000000;
int dp[15][500],num[500],n,m,cnt,map[15];
bool ok(int x)
{
    if(x&(x<<1))return false;
    return true;
}

void init()
{
    memset(num,0,sizeof(num));
    for(int i=0;i<(1<<n);i++)
    {
        if(ok(i))
        num[cnt++]=i;
    }
}

int main()
{
	int i,j,k,t;
    while(scanf("%d%d",&m,&n)!=EOF)
    {
        for(i=0;i<m;i++)
         for( j=0;j<n;j++)
         {
            scanf("%d",&t);
            if(t==0)
			map[i]=map[i]|(1<<j);
         }
         cnt=0;
         init();
         for(i=0;i<cnt;i++)
         if(!(num[i]&map[0]))
         dp[0][i]=1;
         for(i=1;i<m;i++)
         {
            for(j=0;j<cnt;j++)
            {
                if(map[i-1]&num[j])
				continue;
                for(k=0;k<cnt;k++)
                {
                    if(map[i]&num[k])continue;
                    if(num[k]&num[j]) continue;
                    dp[i][k]=(dp[i][k]+dp[i-1][j])%MOD;
                }
            }
         }
         int ans=0;
         for(int i=0;i<cnt;i++)
         ans=(ans+dp[m-1][i])%MOD;
         printf("%d\n",ans);
    }
}