Fliptile POJ - 3279 开关问题

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

思路：当一个格子被翻转偶数次与原来的相同，所以多次翻转是无用的；

          每一行的翻转必然会影响到下一行，可知第一行的翻转状态数有2^m,枚举第一行的翻转状态，下一行受上一行的影响，如果正上方是1，则翻转此格。保证前n-1行都为0；

最后只需要判断最后一行是否全为0，就这一说明这种状态是否可行。。

#include
#include
#include
using namespace std;
const int M=16;
int n,m,cnt;
int maps[M][M],ans[M][M],a[M][M];
int dx[4]={1,-1,0,0};
int dy[4]={0,0,-1,1};
bool check(int x,int y)
{
    if(x<0||x>n||y<0||y>m)
        return 0;
    return 1;
}
void flip(int x,int y)
{
    cnt++;
    a[x][y]=!a[x][y];
    ans[x][y]=1;
    for(int i=0;i<4;i++)
    {
        int nx=x+dx[i],ny=y+dy[i];
        if(check(nx,ny))
            a[nx][ny]=!a[nx][ny];
    }
}
bool solve(int k)
{
    memcpy(a,maps,sizeof(a));//内存拷贝函数
    for(int i=0;icnt)
            {
                ans1=cnt;f=i;
            }
        }
        memset(ans,0,sizeof(ans));
        if(f>=0)
        {
            solve(f);
            for(int i=0;i