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

C++版本

DFS

#include <iostream>
#include <string>
#include <cstring>
using namespace std;
 
int Map[20][20],cal[20][20],out[20][20];
int n,m;
int dir[5][2] = {{0,0},{0,1},{0,-1},{1,0},{-1,0}};
 
int fuc(int x,int y){       //(x,y)的状态由本身的黑白 + 周围五个的翻转状态决定
    int temp = Map[x][y];
 
    for(int i = 0;i < 5;i ++){
        int xi = x+dir[i][0];
        int yi = y+dir[i][1];
 
        if(xi < 1 || xi > n || yi < 1 || yi > m)    continue;
        temp += cal[xi][yi];
    }
    return temp%2;
}
int dfs(){
    for(int i = 2;i <= n;i ++)
        for(int j = 1;j <= m;j ++)
            if(fuc(i-1,j))      //如果上方为黑色，必须要翻转
                cal[i][j] = 1;
 
    for(int i = 1;i <= m;i ++)      //最后一行全白
        if(fuc(n,i))
            return -1;
 
    int res = 0;
    for(int i = 1;i <= n;i ++)
        for(int j = 1;j <= m;j ++)
            res += cal[i][j];
    return res;
}
 
int main()
{
    while(cin>>n>>m){
        for(int i = 1;i <= n;i ++)
            for(int j = 1;j <= m;j ++)
                cin>>Map[i][j];
 
        int flag = 0;
        int ans = 0x3f3f3f3f;
        for(int i = 0;i < 1<<m;i ++){       //第一行 1<<m种状态，二进制从0开始，字典序从小到大
            memset(cal,0,sizeof(cal));
 
            for(int j = 1;j <= m;j ++)      //利用二进制枚举第一行所有的情况
                cal[1][m-j+1] = i>>(j-1) & 1;
            int cont = dfs();
            if(cont >= 0 && cont < ans){        //翻转次数最少
                flag = 1;
                ans = cont;
                memcpy(out,cal,sizeof(cal));
            }
        }
        if(!flag)   cout<<"IMPOSSIBLE"<<endl;
        else{
            for(int i = 1;i <= n;i ++){
                for(int j = 1;j <= m;j ++){
                    if(j != 1)  cout<<" ";
                    cout<<out[i][j];
                }
                cout<<endl;
            }
        }
    }
    return 0;
}