[原创]POJ 3254 Corn Fields [状压DP]【动态规划】

2016-08-03 13:38:04 Tabris_ 阅读数：219

博客爬取于2020-06-14 22:44:01
以下为正文

版权声明：本文为Tabris原创文章，未经博主允许不得私自转载。
https://blog.csdn.net/qq_33184171/article/details/52102801

题目链接 : http://poj.org/problem?id=3254
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Corn Fields
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 12927 Accepted: 6766
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.

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题目大意 : 就是一个n*m的地方 1代表有草 0 代表没草
现在要往这里放一些牛 牛只能放在有草的地方 而且 两个牛不能相邻 问有多少种方法

解题思路 : n*m不大的这种一猜就是状压(一般<=20) 然后每一行与每一行有着状态的转移 于是想到DP 解决

首先吧每一行的草的状态用一个数存起来 就是这样

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int x;
      for(int i=1; i<=n; i++)
      {
          a[i]=0;
          for(int j=0; j<m; j++)
          {
              scanf("%d",&x);
              a[i]=(a[i]<<1)+x;
          }
          //printf("%d",a[i]);
      }

有了这些状态好办了 直接DP

DP 的时候主要判断的就是3项
1.左右有没有挨着的 status&(status<<1)
2.是不是只在有草的地方放牛 (草地的状态&想放牛的状态)==想放牛的状态
3.判断相邻两行的状态有没有相邻的(这时候相邻就是上下相邻) 直接 上&下 就行了

然后就是DP一下就好了 本人做这题的时候也是刚学DP 博客写的很…挫…

附本题代码
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# include<cstdio>
# include<iostream>
# include<algorithm>
# include<cmath>
# include<cstdlib>
# include<cstring>
# include<iostream>
using namespace std;

typedef long long LL;

int dp[13][(1<<12)+10];
int a[13];
int n,m;
const int MOD = 1e8;

int main()
{

    while(~scanf("%d%d",&n,&m))
    {
        memset(dp,0,sizeof(dp));

        int x;
        for(int i=1; i<=n; i++)
        {
            a[i]=0;
            for(int j=0; j<m; j++)
            {
                scanf("%d",&x);
                a[i]=(a[i]<<1)+x;
            }
            //printf("%d",a[i]);
        }

        dp[0][0]=1;
        for(int i=1; i<=n; i++)
        {
            for(int j=0; j<(1<<m); j++)
            {
                if((j&(j<<1))!=0||(a[i]&j)!=j) continue;//判断有没有临近的
                for(int k=0; k<(1<<m); k++)
                {
                    if((j&k)!=0) continue;
                    dp[i][j]=(dp[i][j]+dp[i-1][k]+MOD)%MOD;
                }
            }
        }

        int res=0;
        for(int i=0; i<(1<<m); i++)
            res=(res+dp[n][i]+MOD)%MOD;

        printf("%d\n",res);
    }
    return 0;
}