POJ2965 The Pilots Brothers' refrigerator

题目链接：http://poj.org/problem?id=2965

大致题意：这题和POJ1753 Flip GAME差不多，就是在这里反转时是一整行一整列都要变化，而不像1753题只是相邻的上下左右要变化，具体见https://blog.youkuaiyun.com/Fitz1318/article/details/86146329

解题思路：具体见https://blog.youkuaiyun.com/Fitz1318/article/details/86146329

AC代码：

#include <iostream>
using namespace std;

bool a[5][5] = {0};
int ri[16],cj[16];
bool flag;
int deep;
int step;

int judge()
{
    for(int i=1; i<5; i++){
        for(int j=1; j<5; j++){
            if(a[i][j] != 1)
                return 0;
        }
    }
    return 1;
}

void flip(int x,int y)
{
    for(int i=1; i<5; i++){
        a[x][i] = !a[x][i];
    }
    for(int j=1; j<5; j++){
        a[j][y] = !a[j][y];
    }
    a[x][y] =!a[x][y];
}

void dfs(int x,int y,int deep)
{
    if(deep == step){
        flag = judge();
        return;
    }
    if(flag||x==5)
        return;

    //翻棋
    flip(x,y);
    ri[deep] = x;
    cj[deep] = y;
    if(y<4)
        dfs(x,y+1,deep+1);
    else
        dfs(x+1,1,deep+1);

    //不符合则翻转回来
    flip(x,y);
    if(y<4)
        dfs(x,y+1,deep);
    else
        dfs(x+1,1,deep);
    return;
}
int main()
{
    char temp;
    int i,j;
    for(i=1; i<5; i++){
        for(j=1; j<5;j++){
            cin >> temp;
            if(temp == '-')
                a[i][j] = 1;
        }
    }
    for(step=0; step<=16; step++){
        dfs(1,1,0);
        if(flag)
            break;
    }
    cout << step << endl;
    for(int i=0; i<step; i++){
      cout << ri[i] <<' '<<cj[i]<< endl;
    }
    return 0;
}