[CareerCup] 18.11 Maximum Subsquare 最大子方形

18.11 Imagine you have a square matrix, where each cell (pixel) is either black or white. Design an algorithm to find the maximum subsquare such that all four borders are filled with black pixels.

LeetCode上的原题，请参见我之前的解法Maximal Square。书上给了两种解法，但是比较长：

解法一：

class Subsquare {
public:
    int row, col, size;
    Subsquare(int r, int c, int sz): row(r), col(c), size(sz) {}
    void print() {
        cout << "(" << row << ", " << col << ", " << size << ")" << endl;
    }
};

bool is_square(vector<vector<int>> &matrix, int row, int col, int size) {
    for (int j = ; j < size; ++j) {
        if (matrix[row][col + j] == ) return false;
        if (matrix[row + size - ][col + j] == ) return false;
    }
    for (int i = ; i < size - ; ++i) {
        if (matrix[row + i][col] == ) return false;
        if (matrix[row + i][col + size - ] == ) return false;
    }
    return true;
}

Subsquare* find_square_with_size(vector<vector<int>> &matrix, int squareSize) {
    int cnt = matrix.size() - squareSize + ;
    for (int row = ; row < cnt; ++row) {
        for (int col = ; col < cnt; ++col) {
            if (is_square(matrix, row, col, squareSize)) {
                return new Subsquare(row, col, squareSize);
            }
        }
    }
    return NULL;
}

Subsquare* find_square(vector<vector<int>> &matrix) {
    for (int i = matrix.size(); i >= ; --i) {
        Subsquare *square = find_square_with_size(matrix, i);
        if (square) return square;
    }
    return NULL;
}

解法二：

class Subsquare {
public:
    int row, col, size;
    Subsquare(int r, int c, int sz): row(r), col(c), size(sz) {}
    void print() {
        cout << "(" << row << ", " << col << ", " << size << ")" << endl;
    }
};

class SquareCell {
public:
    int zerosRight = , zerosBelow = ;
    SquareCell(int right, int below): zerosRight(right), zerosBelow(below){}
    void setZerosRight(int right) {
        zerosRight = right;
    }
    void setZerosBelow(int below) {
        zerosBelow = below;
    }
};

bool is_square(vector<vector<SquareCell*>> &matrix, int row, int col, int size) {
    SquareCell *topLeft = matrix[row][col];
    SquareCell *topRight = matrix[row][col + size - ];
    SquareCell *bottomRight = matrix[row + size - ][col];
    if (topLeft->zerosRight < size) return false;
    if (topLeft->zerosBelow < size) return false;
    if (topRight->zerosBelow < size) return false;
    if (bottomRight->zerosRight < size) return false;
    return true;
}

vector<vector<SquareCell*>> process_square(vector<vector<int>> &matrix) {
    vector<vector<SquareCell*>> res(matrix.size(), vector<SquareCell*>(matrix.size()));
    for (int r = matrix.size() - ; r >= ; --r) {
        for (int c = matrix.size() - ; c >= ; --c) {
            int rightZeros = , belowZeros = ;
            if (matrix[r][c] == ) {
                ++rightZeros;
                ++belowZeros;
                if (c +  < matrix.size()) {
                    SquareCell *pre = res[r][c + ];
                    rightZeros += pre->zerosRight;
                }
                if (r +  < matrix.size()) {
                    SquareCell *pre = res[r + ][c];
                    belowZeros += pre->zerosBelow;
                }
            }
            res[r][c] = new SquareCell(rightZeros, belowZeros);
        }
    }
    return res;
}

Subsquare* find_square_with_size(vector<vector<SquareCell*>> &processed, int square_size) {
    int cnt = processed.size() - square_size + ;
    for (int row = ; row < cnt; ++row) {
        for (int col = ; col < cnt; ++col) {
            if (is_square(processed, row, col, square_size)) {
                return new Subsquare(row, col, square_size);
            }
        }
    }
    return NULL;
}

Subsquare* find_square(vector<vector<int>> &matrix) {
    vector<vector<SquareCell*>> processed = process_square(matrix);
    // cout << "here" << endl;
    for (int i = matrix.size(); i >= ; --i) {
        Subsquare *square = find_square_with_size(processed, i);
        if (square) return square;
    }
    return NULL;
}

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