D - Fliptile

Farmer John knows that an intellectually satisfied cow is a happy cow who will give more milk. He has arranged a brainy activity for cows in which they manipulate an M × N grid (1 ≤ M ≤ 15; 1 ≤ N ≤ 15) of square tiles, each of which is colored black on one side and white on the other side.

As one would guess, when a single white tile is flipped, it changes to black; when a single black tile is flipped, it changes to white. The cows are rewarded when they flip the tiles so that each tile has the white side face up. However, the cows have rather large hooves and when they try to flip a certain tile, they also flip all the adjacent tiles (tiles that share a full edge with the flipped tile). Since the flips are tiring, the cows want to minimize the number of flips they have to make.

Help the cows determine the minimum number of flips required, and the locations to flip to achieve that minimum. If there are multiple ways to achieve the task with the minimum amount of flips, return the one with the least lexicographical ordering in the output when considered as a string. If the task is impossible, print one line with the word “IMPOSSIBLE”.

Input
Line 1: Two space-separated integers: M and N
Lines 2… M+1: Line i+1 describes the colors (left to right) of row i of the grid with N space-separated integers which are 1 for black and 0 for white
Output
Lines 1… M: Each line contains N space-separated integers, each specifying how many times to flip that particular location.
Sample Input
4 4
1 0 0 1
0 1 1 0
0 1 1 0
1 0 0 1
Sample Output
0 0 0 0
1 0 0 1
1 0 0 1
0 0 0 0
题意大概就是农夫要检测奶牛的智商，让奶牛反转棋盘，使得棋盘最后都是白色朝上，反转的规则是反转一个棋子就会使得该棋子上下左右的棋子都反转，求问最后反转次数最少且字典序最小的反转表，如果不能实现输出IMPOSSIBLE
思路是这样的，我们按照字典序枚举第一行的2^N种情况，
for (int i = 0; i < 1 << N; i++)//进行2^N次循环
{
memset(flip, 0, sizeof(flip));
for (int j = 0; j < N; j++)
flip[0][N - j - 1] = i >> j & 1;
//0000 0001 0010 0011 0100 这个顺序反转，1代表反转，0代表不反转
……
（利用位运算实现，我觉得想出这个方法的人真的很厉害）
枚举出第一行的反转情况后，从第二行开始，我们只要以将前一行的黑棋变成白色为目的，那么就可以保证除了最后一行，前面的棋子都为白色。
那么最后检验最后一行是全否为白子即可确定当前枚举是否成立。
代码如下：//如有错误欢迎指正

#include
#include
using namespace std;
const int dx[5] = { -1,0,0,0,1 };
const int dy[5] = { 0,-1,0,1,0 };
int M, N;
int tile[20][20];
int opt[20][20];//保存最优解
int flip[20][20];//保存中间结果
//查询(x,y)的颜色
int get(int x, int y)
{
	int c = tile[x][y];
	for (int d = 0; d < 5; d++)
	{
		int x2 = x + dx[d], y2 = y + dy[d];
		if (x2 >= 0 && x2 < M&&y2 >= 0 && y2 < N)
			c += flip[x2][y2];//他前后左右被反转的总和和他自己本身的颜色的和
	}
	return c % 2;//对二取余即能确定他的颜色
}
int cal()
{
	for(int i=1;i> j & 1;//以0000 0001 0010 0011这个规律
		int num = cal();
		if (num >= 0 && (ans < 0 || ans>num))
		{
			ans = num;
			memcpy(opt, flip, sizeof(flip));
		}
	}
	if (ans < 0)
		cout << "IMPOSSIBLE\n";
	else {
		for (int i = 0; i < M; i++) {
			for (int j = 0; j < N; j++) {
				cout << opt[i][j] << (j + 1 == N ? '\n' : ' ');
			}
		}
	}
}
int main()
{
	while (cin >> M >> N)
	{
		memset(tile, 0, sizeof(tile));
		for (int i = 0; i < M; i++)
			for (int j = 0; j < N; j++)
				cin >> tile[i][j];
		solve();
	}
	return 0;
}