Description
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
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
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：有一个n*m的棋盘，0表示白色，1表示黑色。每次可以翻转当前位置，它的上下左右四个位置也会被相应翻转。问最少翻转多少次会使所有棋面显示为白色，并给出需要翻转的位置，0表示不翻转，1表示翻转。
思路：可以利用第一层的 2^n 种状态来进行枚举。依次翻转到最后一行，如果最后一行均为白色，那么这种第一层状态可以达到效果。记录下来，最后取最少翻转次数的结果输出。具体实现细节和相关解释在代码注释中给出。
ps:(学到了利用二进制进行枚举的方便之处)QAQ~
代码如下：
#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;
}