POJ 3254 Corn Fields【状态压缩DP】

最新推荐文章于 2023-08-08 14:59:34 发布

转载最新推荐文章于 2023-08-08 14:59:34 发布 · 157 阅读

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原文链接：http://www.cnblogs.com/dream-wind/archive/2012/03/30/2426089.html

本文解决了一个经典的农场种植问题，即如何在有限且不连续的肥沃土地上种植玉米，同时确保相邻地块不会被同时选择。文章提供了一段C语言代码实现，通过动态规划的方法来计算所有可能的选择方案。

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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

code：

View Code

#include<stdio.h>
#include<string.h>
#define mod 100000000
#define clr(x)memset(x,0,sizeof(x))
int co(int x)
{
if(x<2)
return 1;
if(x&1 && x&2)
return 0;
return co(x>>1);
}
int last[1<<12];
int state[12];
int next[1<<12];
int ss[1<<13];
int main()
{
int top=0,m,n,i,j,k,p,s,res;
while(scanf("%d%d",&m,&n)!=EOF)
    {
for(i=0;i<m;i++)
        {
            state[i]=0;
for(j=0;j<n;j++)
            {
                scanf("%d",&p);
                state[i]=(state[i]<<1)+1-p;
            }
        }
        s=(1<<n)-1;
for(i=0;i<s;i++)
if(co(i)==1)
                ss[i]=1;
else ss[i]=0;
for(i=0;i<s;i++)
        {
if(ss[i]==0||i&state[0])
                last[i]=0;
else last[i]=1;
        }
for(i=1;i<m;i++)
        {
for(j=0;j<s;j++)
            {
                next[j]=0;
if(ss[j]==0||j&state[i])
continue;
for(k=0;k<s;k++)
                {
if(j&k) continue;
                    next[j]=(next[j]+last[k])%mod;
                }
            }
for(j=0;j<s;j++)
                last[j]=next[j];
        }
        res=0;
for(i=0;i<s;i++)
            res=(res+last[i])%mod;
        printf("%d\n",res);
    }
return 0;
}

转载于:https://www.cnblogs.com/dream-wind/archive/2012/03/30/2426089.html