Corn Fields - POJ 3254 状压dp-优快云博客

本文链接：https://blog.youkuaiyun.com/u014733623/article/details/37529231

本文介绍了一个关于在矩形牧场上种植玉米的问题，使用状态压缩动态规划（状压DP）的方法来求解。考虑了土地肥沃性及避免相邻种植的约束条件，通过代码实现展示了具体的解题过程。

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Corn Fields

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 6800		Accepted: 3614

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

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：牛只能在肥沃的土地上吃草，并且两头牛不能上下左右挨着，问你可以有多少种吃草方案。

思路：状压dp。先把一行内不挨着的情况找出来，一共是980多种，然后第一行先判断这些情况存在的。然后后面的行，每种情况先判断这种情况在该行能不能满足，再判断与上一行是否有挨着的。

AC代码如下：

#include<cstdio>
#include<cstring>
using namespace std;
int row[20],state[1010];
int dp[20][1010],mod=100000000,ans;
int main()
{ int m,n,i,j,k,num=0;
  scanf("%d%d",&m,&n);
  k=1<<n;
  for(i=0;i<k;i++)
   if((i&(i<<1))==0)
    state[num++]=i;
  for(i=0;i<m;i++)
   for(j=n-1;j>=0;j--)
   { scanf("%d",&k);
     row[i]+=k<<j;
   }
  for(i=0;i<num;i++)
   if((row[0]&state[i])==state[i])
    dp[0][i]=1;
  for(i=1;i<m;i++)
   for(j=0;j<num;j++)
   { if((row[i]&state[j])!=state[j])
      continue;
     for(k=0;k<num;k++)
      if(dp[i-1][k] && ((state[k] & state[j])==0))
       dp[i][j]=(dp[i][j]+dp[i-1][k])%mod;
   }
  ans=0;
  for(i=0;i<num;i++)
   ans=(ans+dp[m-1][i])%mod;
  printf("%d\n",ans);
}