POJ 3254 Corn Fields （状压）-优快云博客

本文链接：https://blog.youkuaiyun.com/u014737310/article/details/44173653

本文探讨了农夫约翰在种植玉米时如何在受限制的地块中做出选择，确保每块地与其他地块不相邻。通过详细分析地块布局和种植条件，文章提供了多种选择方案的数量计算，包括不种植任何地块的情况。这为农民在资源有限的情况下提供了灵活的种植策略。

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Language:

Corn Fields

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 8704		Accepted: 4641

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.

Source

USACO 2006 November Gold

下面链接题解很清楚

参考链接 : http://www.cnblogs.com/void/articles/2072601.html

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>


#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define eps 1e-8
typedef __int64 ll;

#define fre(i,a,b)  for(i = a; i <b; i++)
#define fro(i,a,b)  fro(i = b; i >=a; i--)
#define mem(t, v)   memset ((t) , v, sizeof(t))

#define sfs(a)      scanf("%s", a);
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)

#define pf          printf
#define bug         pf("Hi\n")

using namespace std;

#define INF 0x3f3f3f3f
#define N 1<<12

#define mod 100000000

int dp[12][N],h[N],type[N];
int num,n,m;


bool judge(int x)
{
	if(x&(x<<1)) return false;
	return true;
}

void inint()
{

	int len=1<<m,i;

    fre(i,0,len)
       if(judge(i))
		type[num++]=i;

}

int main()
{
	int i,j,t;
	while(~sff(n,m))
	{
		mem(h,0);
		fre(i,0,n)
		  fre(j,0,m)
		  {
		  	int x;
		  	sf(x);
		  	if(x==0) h[i]|=1<<j;
		  }

		num=0;
		inint();

	   mem(dp,0);
       fre(i,0,num)
        if((h[0]&type[i])==0)
		{
			dp[0][i]=1;
		}

	   fre(i,1,n)
	     fre(j,0,num)
	     {
	     	if(h[i]&type[j]) continue;

            fre(t,0,num)
            {
            	if((h[i-1]&type[t])||(type[j]&type[t]))  continue;
            	dp[i][j]=(dp[i][j]+dp[i-1][t])%mod;

            }
	     }

	int ans=0;

	fre(i,0,num)
	  ans=(ans+dp[n-1][i])%mod;
	pf("%d\n",ans);

	}

	return 0;
}