zoukankan      html  css  js  c++  java
  • [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 = 0; j < size; ++j) {
            if (matrix[row][col + j] == 1) return false;
            if (matrix[row + size - 1][col + j] == 1) return false;
        }
        for (int i = 1; i < size - 1; ++i) {
            if (matrix[row + i][col] == 1) return false;
            if (matrix[row + i][col + size - 1] == 1) return false;
        }
        return true;
    }
    
    Subsquare* find_square_with_size(vector<vector<int>> &matrix, int squareSize) {
        int cnt = matrix.size() - squareSize + 1;
        for (int row = 0; row < cnt; ++row) {
            for (int col = 0; 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 >= 1; --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 = 0, zerosBelow = 0;
        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 - 1];
        SquareCell *bottomRight = matrix[row + size - 1][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() - 1; r >= 0; --r) {
            for (int c = matrix.size() - 1; c >= 0; --c) {
                int rightZeros = 0, belowZeros = 0;
                if (matrix[r][c] == 0) {
                    ++rightZeros;
                    ++belowZeros;
                    if (c + 1 < matrix.size()) {
                        SquareCell *pre = res[r][c + 1];
                        rightZeros += pre->zerosRight;
                    }
                    if (r + 1 < matrix.size()) {
                        SquareCell *pre = res[r + 1][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 + 1;
        for (int row = 0; row < cnt; ++row) {
            for (int col = 0; 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 >= 1; --i) {
            Subsquare *square = find_square_with_size(processed, i);
            if (square) return square;
        }
        return NULL;
    }

    CareerCup All in One 题目汇总

  • 相关阅读:
    SOJ 1035 DNA matching
    SOJ 1027 MJ,Nowhere to Hide
    SOJ 1021 Couples
    SOJ 1020 Big Integer
    C#中正则表达式的简单使用
    根据HTTP header收集客户端相关信息 --- tornado demo
    IIS7配置Gzip压缩
    网站性能工具Yslow的使用方法
    配置ETags
    细说ETags以及如何在 IIS6和 IIS7下取消ETags
  • 原文地址:https://www.cnblogs.com/grandyang/p/5484008.html
Copyright © 2011-2022 走看看