状态压缩DP POJ 3254 Corn Fields（玉米地）

题目：

 

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

 

 

 

这个题目，因为递推比较简单，而且m和n也比较小，所以总体比较简单。

代码：

#include
#include
using namespace std;

int list[13];
int r[13][4096];

bool ok(int n)		//一行里面不能有相邻的2个
{
	for (int i = 3; i <= 3072; i *= 2)if ((n&i) == i)return false;
	return true;
}

int main()
{
	int m, n;
	int a;
	cin >> m >> n;
	memset(list, 0, sizeof(list));
	for (int i = 1; i <= m; i++)		//输入
	{
		for (int j = 0; j < n; j++)
		{
			cin >> a;
			if (a)list[i] += (1 << j);
		}
	}
	memset(r, 0, sizeof(r));
	for (int i = 1; i <= m; i++)for (int j = 0; j < (1 << n); j++)		//求解
	if (ok(j) && (j&list[i]) == j)		//单就这一行来说是满足的
	{
		if (i == 1)r[1][j] = 1;
		else for (int k = 0; k < (1 << n); k++)
			if ((k&j) == 0)r[i][j] = (r[i - 1][k] + r[i][j]) % 100000000;
	}
	int s = 0;
	for (int i = 0; i < (1 << n); i++)s = (s + r[m][i]) % 100000000;	//对最后一行求和
	cout << s;
	return 0;
}