[Locked] Smallest Rectangle Enclosing Black Pixels

An image is represented by a binary matrix with 0 as a white pixel and 1 as a black pixel. The black pixels are connected, i.e., there is only one black region. Pixels are connected horizontally and vertically. Given the location (x, y) of one of the black pixels, return the area of the smallest (axis-aligned) rectangle that encloses all black pixels.

For example, given the following image:

[
  "0010",
  "0110",
  "0100"
]

and x = 0, y = 2,

Return 6.

分析：

　　典型Flood Fill泛洪算法的应用，从一个点扩展到整个区域。

代码：

class Solution {
private:
    int xmin = INT_MAX, xmax = INT_MIN, ymin = INT_MAX, ymax = INT_MIN;
public:
    inline void adjust(int i, int j) {
        xmin = min(xmin, i);
        xmax = max(xmax, i);
        ymin = min(ymin, j);
        ymax = max(ymax, j);
        return;
    }
    void dfs(vector<vector<int> > &matrix, int i, int j, vector<vector<bool> > visited) {
        if(!matrix[i][j] || visited[i][j])
            return;
        //调整最大边缘点
        adjust(i, j);
        visited[i][j] = true;
        dfs(matrix, i + 1, j, visited);
        dfs(matrix, i - 1, j, visited);
        dfs(matrix, i, j + 1, visited);
        dfs(matrix, i, j - 1, visited);
        visited[i][j] = false;
        return;
    }
    int minArea(vector<vector<int> > &matrix, int x, int y) {
        //设立岗哨，避免递归内部为了判断边界而产生大量开销
        matrix.insert(matrix.begin(), vector<int> (matrix[0].size(), 0));
        matrix.push_back(vector<int> (matrix[0].size(), 0));
        for(auto &m : matrix) {
            m.insert(m.begin(), 0);
            m.push_back(0);
        }
        //避免重复访问
        vector<vector<bool> > visited(matrix.size(), vector<bool> (matrix[0].size(), false));
        dfs(matrix, x + 1, y + 1, visited);
        return (ymax - ymin + 1) * (xmax - xmin + 1);
    }
};