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cclassshape
- 声明一个基类Shape(点), 在此基础上派生出Rectangle(长方形)和Circle(圆),这三个类都有GetArea()函数计算对象的面积,构造函数,析构函数等有关函数。再使用Rectangle类创建一个派生类Square(正方形)。并设计创建各种类的对象,调用所有函数。设计函数f(Shape &a)能对不同对象的实参调用计算打印出对象的面积。 -statement a base class Shape (points) On this basis -- for Rectangle
EXTENDS
- C++入门实例,对继承与组合的初步应用。可以让用户选择在指定位置打印出不同的图形,如正立等腰三角形,倒立等腰三角形,矩形,正方形、菱形和多边形。图形由随机选择的*或者#组成。 请为每种图形设计一个类,这些类之间的关系可能是继承或者组合。 每个类需要能保存该图形对象的参数,改变参数的方法以及在屏幕上用字符打印该图形的方法。 程序根据用户的选择产生一个新图形的对象,为该对象设置参数并打印。 -C++ Getting Started ex
JudgePointInRect
- 根据矩形左边中点、方向及长宽求矩形各点坐标;判断平面上一点是否在该平面上某一不规则矩形内-According to the midpoint of the rectangle on the left, direction and length and width of a rectangular coordinate points determine whether the plane a little irregular in the rectangular plane within a
juxingxiangcheng
- 矩形相乘 很好的 和大家分享 欢迎大家下载哦-Rectangle multiply very good to share welcome to download Oh
maze
- 采用C++写的找迷宫的算法,可以找到矩形迷宫中的路径,并且可以处理一些迷宫中遇到的阻碍-Written using C++ algorithm to find the maze, the maze can be found in the path of the rectangle, and can handle some of the obstacles encountered in the maze
seg
- 线段树经典应用,求合并矩形的周长。IOI经典题目Picture。-Classic application of the segment tree, find the perimeter of the rectangle consolidation.
1
- 在一个由M*N个小正方形组成的矩形上,依次用两个相等的小正方形大小的矩形(此处所谓相邻是指两个正方形有一条公共边)不重复的覆盖该矩形区域。请构造出该问题的模型,设计算法求解所有可能的覆盖。-In a M* N small squares on the rectangle, followed by two small squares of equal size of the rectangle (here is the so-called two adjacent squares have a c
stereo-matching
- 关于自适应窗口算法的描述,很详细,是最经典的自适应窗口算法,值得学习-We present a method to select an appropriate window by evaluating the local variation of the intensity and the disparity. We employ a statistical model of the disparity distribution within the window. This mod
CPP
- C++绘图代码,绘制矩形,圆,三角形,直线,折线等较简单的平面图形-C++ drawing code to draw a rectangle, circle, triangle, straight lines, polylines, and other relatively simple plane figure
rec_jt_lx_cl
- 使用c++语言在vc环境下编写的可画出矩形、箭头、菱形和圆形的小程序。-A programm of drawing a rectangle, an arrow, a diamond and a circular using c++ language in vc environment.
DP
- 动态规划问题。给定一个整数的二维数组,由其中若干邻近元素构成的矩形称为子数组。 编程计算所有子数组元素之和的最大值。输入为整数n,及n^2个数,输出为所有子数组元素之和的最大值-Dynamic programming problem.Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1*1 or greater l
Nested-Matrix
- 有n个矩形,选出尽可能多的矩形排成一行,使得除最后一个外,每一个矩形都可以嵌套在下一个矩形内。-There are n rectangles, rectangles selected as many aligned such that except the last, are each rectangle nested within the next rectangle.
rectangle_c++
- rectangle in c++ prgramming uni
Largest-Rectangle-in-Histogram
- LeetCode面试题目,求最大矩形的情况,动态规划-LeetCode interview questions, find the largest rectangular, dynamic programming
Maze
- 为了测试某种药物对小白鼠方向感的影响,生物学家在实验室做了一个矩形迷宫,入口和出口都确定为唯一的,且分布在矩形的不同边上。现在让你算出小白鼠最短需要走多少步,才可以从入口走到出口。-In order to test the impact of a drug on mice sense of direction, a biologist at the lab made a rectangular maze, inlet and outlet are determined to be unique
landmine
- THUDSA2019Fall。利用数据结构栈,实现直方图内最大矩形的查找,根据最大矩形确定地雷数目(THUDSA2019fall. Using stack to find the max area in a rectangle.)
