題目：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
解：因爲下面一行每次只能影響上一行的一塊，要全部翻好則第一行確定時，就已經確定翻多少
只要搜索第一行所有的情況就可以
只有翻與不翻類似二進制0和1，有2的n次方種情況，一直求餘二則爲求二進制的方法。
代碼:
#include<iostream>
using namespace std;
int a[20][20];
int b[20][20];
int c[20][20];
int d[20][20];
int account;
int pint;
int lock;
void remark(int x,int y)
{
pint = account;
for (int i = 1; i <= x; i++)
{
for (int j = 1; j <= y; j++)
{
d[i][j] = c[i][j];
}
}
}
void turn(int x, int y)
{
c[x][y] = 1;
account++;
b[x][y] = 1 - b[x][y];
b[x - 1][y] = 1 - b[x - 1][y];
b[x + 1][y] = 1 - b[x + 1][y];
b[x][y - 1] = 1 - b[x][y - 1];
b[x][y + 1] = 1 - b[x][y + 1];
}
int main()
{
int m, n;
while (cin >> m >> n)
{
for (int i = 1; i <= m; i++)
{
for (int j = 1; j <= n; j++)
{
cin >> a[i][j];
}
}
int t = 1 << n;//第一行的情況種類數，向左位運算類似於乘2(計算每種第一行的情況所需次數)
for (int i = 0; i < t; i++)
{
memset(c, 0, sizeof(c));
for (int i = 1; i <= m; i++)
{
for (int j = 1; j <= n; j++)
{
b[i][j] = a[i][j];
}
}
int pm = t;
account = 0;
pint = 99999;
for (int i = 1; i <= pm; i++) //翻轉第一行
{
if (pm % 2 == 1)
{
turn(1, i);
}
pm /= 2;
}
for (int i = 2; i <= m; i++)
{
for (int j = 1; j <= n; j++)
{
if (b[i - 1][j] == 1)
{
turn(i, j);
}
}
}
for (int j = 1; j <= n; j++)
{
if (b[m][j] == 1)
{
break;
}
if (j == n)
{
lock = 1; //只要成功一次不會是not impossible
if (account < pint)
{
remark(m, n);
}
}
}
}
if (lock == 0)
{
cout << "IMPOSSIBLE" << endl;
return 0;
}
for (int i = 1; i <= m; i++)
{
for (int j = 1; j <= n; j++)
{
if (j != 1)
{
cout << " ";
}
cout << d[i][j];
if (j == n)
{
cout << endl;
}
}
}
}
return 0;
}