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)的狀態(tài)由本身的黑白 + 周圍五個的翻轉(zhuǎn)狀態(tài)決定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)) //如果上方為黑色，必須要翻轉(zhuǎn)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種狀態(tài)，二進制從0開始，字典序從小到大memset(cal,0,sizeof(cal));for(int j = 1;j <= m;j ++) //利用二進制枚舉第一行所有的情況cal[1][m-j+1] = i>>(j-1) & 1;int cont = dfs();if(cont >= 0 && cont < ans){ //翻轉(zhuǎn)次數(shù)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; }

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