poj 3254 Corn Fields

最新推荐文章于 2022-02-03 21:36:08 发布

brucehb

最新推荐文章于 2022-02-03 21:36:08 发布

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分类专栏：状态压缩dp

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博客围绕农场主John的种植问题展开，他有M×N的牧场，部分方格肥沃可种玉米，且所选方格不能相邻。需计算种植方格的选择方案数。采用状态压缩dp方法，dp[i][j]表示第i行状态为j时的所有可能，通过特定条件判断状态并更新dp值。

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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

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.

思路：

状态压缩dp。dp[i][j]表示第i行状态为j时的所有可能。

判断一个状态没有相邻1时用：state & (state << 1) == 0。

从上到下逐行扫描，每次扫描该行的一个状态，然后更新在这个状态下的下面一行的所有状态的dp。

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;

typedef long long ll;
const int MAXN = 13;
const int MOD = 1e9;
ll dp[MAXN][1 << MAXN];
int n, m;

int infertile[MAXN];

int main()
{
	cin >> m >> n;
    	
	int a;
	memset(dp, 0, sizeof(dp));

	for (int i = 1; i <= m; i++)
	{
		int temp = 0;
		for (int j = 0; j < n; j++)
		{
		    cin >> a;
			if (a == 0)
			{
				temp |= (1 << j);
			}
		}
		infertile[i] = temp;
	}
	infertile[0] = 0;
	dp[0][0] = 1;

	for (int i = 0; i < m; i++)
	{
		for (int j = 0; j < (1 << n); j++)
		{
			if (((j & infertile[i]) == 0) && ((j & (j << 1)) == 0))
			{
				for (int k = 0; k < (1 << n); k++)
				{
					if (((k & infertile[i+1]) == 0) && ((k & (k << 1)) == 0) && ((j & k) == 0))
					{
						dp[i+1][k] = (dp[i+1][k] + dp[i][j]) % MOD;
					}
				}
			}

		}
	}

	ll result = 0;
	for (int i = 0; i < (1 << n); i++)
	{
		result = (result + dp[m][i]) % MOD;
	}

	cout << result << endl;

	return 0;
}