1284. Minimum Number of Flips to Convert Binary Matrix to Zero Matrix

看了答案，加上每个点只能走一次这个前提，就可以进行dfs暴力搜索了。【因为一个点走两次==这个点没走，无论中间隔了多少个点都成立】

class Solution {
public:
    // 如何将一个mats经过若干步骤后翻转成全0的矩阵？？
    // 每次翻转，将改变四周以及当前位置
    
    void flip(vector<vector<int>>& mat, int x, int y){
        int row = mat.size();
        int col = mat[0].size();
        mat[x][y] = 1-mat[x][y];
        if (x-1>=0&&x-1<row&&y>=0&&y<col){
            mat[x-1][y] = 1-mat[x-1][y];
        }
        if (x+1>=0&&x+1<row&&y>=0&&y<col){
            mat[x+1][y] = 1-mat[x+1][y];
        }
        if (x>=0&&x<row&&y-1>=0&&y-1<col){
            mat[x][y-1] = 1-mat[x][y-1];
        }
        if (x>=0&&x<row&&y+1>=0&&y+1<col){
            mat[x][y+1] = 1-mat[x][y+1];
        }
        return;
    }
    
    bool checkzeros(vector<vector<int>>& mat){
        int row = mat.size();
        int col = mat[0].size();
        for (int i=0;i<row;i++){
            for(int j=0;j<col;j++){
                if (mat[i][j]==1){
                    return false;
                }
            }
        }
        return true;
    }
    
    
    int func(vector<vector<int>>& mat, vector<vector<int>>& visit) {
        int row = mat.size();
        int col = mat[0].size();
        if (checkzeros(mat)){return 0;}
        
        int result = 1e9;
        for(int i=0;i<row;i++){
            for(int j=0;j<col;j++){
                //cout<<i<<" "<<j<<endl;
                if (visit[i][j]==0){
                    visit[i][j]=1;
                    flip(mat, i, j);
                    result = min(result, func(mat, visit)+1);
                    flip(mat, i, j);
                    visit[i][j]=0;
                }
            }
        }
        
        return result;
        
    }
    
     int minFlips(vector<vector<int>>& mat) {
        int row = mat.size();
        int col = mat[0].size();
        vector<int> temp(col, 0);
        vector<vector<int>> visit(row, temp);
        int res = func(mat, visit);
        if (res>=1e9){return -1;}
         return res;        
    }
};