Description: 用C++中的MFC编程实现正轴等角割圆柱投影，实现以下要求：
取克拉索夫斯基椭球
(1)制图区域： Bs=0°, BN=25°
LE=105°, LE=125°
(2)经纬线间隔： ΔB=ΔL=5°
(3)制图比例尺： 1:M0=1:1000 000
(4)标准纬线： Bk=±15°
计算经纬网格点的 x, y,m,n, p
-With C++ Of MFC programming is cutting Conformal cylindrical projection axis to achieve the following requirements: check克拉索Malinowski ellipsoid (1) mapping the region: Bs = 0 °, BN = 25 ° LE = 105 °, LE = 125 ° (2) latitude and longitude line spacing: ΔB = ΔL = 5 ° (3) Drawing scale: 1: M0 = 1:1000 000 (4) the standard parallels: Bk = ± 15 ° latitude and longitude grid computing point x , y, m, n, p Platform: |
Size: 40960 |
Author:张建 |
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Description: 在一个划分成网格的操场上，n个士兵散乱地站在网格点上。网格点由整数坐标(x,y)表示。士兵们可以沿网格边上、下、左、右移动一步，但在同一时刻任一网格点上只能有一名士兵。按照军官的命令，士兵们要整齐地列成一个水平队列，即排列成(x,y),(x+1,y),…,(x+n-1,y)。如何选择x 和y的值才能使士兵们以最少的总移动步数排成一列。
实验任务：计算使所有士兵排成一行需要的最少移动步数。-In a grid divided into the playground, n soldiers stood scattered on the grid points. Grid point by the integer coordinates (x, y) express. Soldiers along the edge of the grid, down, left and right moving step, but at the same time any one grid point has only one soldier. In accordance with the officer s command, the soldiers should be neatly列成level queue, that is arranged in (x, y), (x+ 1, y), ..., (x+ N-1, y). How to choose x and y values to make the soldiers with the minimum of the total moves one line up. Experimental tasks: the calculation so that all the soldiers lined up his required minimum moves. Platform: |
Size: 1024 |
Author:林雨 |
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Description: 有一个m*n格的迷宫(表示有m行、n列)，其中有可走的也有不可走的，如果用1表示可以走，0表示不可以走，文件读入这m*n个数据和起始点、结束点(起始点和结束点都是用两个数据来描述的，分别表示这个点的行号和列号)。现在要你编程找出所有可行的道路，要求所走的路中没有重复的点，走时只能是上下左右四个方向。如果一条路都不可行，则输出相应信息(用-l表示无路)。 -A m* n grid of a maze (that has m lines, n out), which may also not go away, if that can go with 1, 0 that can not leave, this document is read into m* n data and the starting point and end point (the starting point and end point are used to describe the two data, respectively, the point that the line number and column number). Programming you have to find all possible roads, the request path does not repeat the point, can only be walking up and down around the four directions. If a road is not feasible, then output the corresponding information (that have no use-l). Platform: |
Size: 1024 |
Author:肖影 |
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Description: 作业名称： 简易的图形编辑器
实现功能：A: 有一些图元（例如：点，线，矩形，园...）
B: 可以在任意位置任意方向作图
C: 填充方式至少有四种（实的，网格，竖线，横线等）
D：可以扩充图元课填充的方法
-Operating Name: simple graphics editor to achieve function: A: there are some graphic elements (such as: point, line, rectangle, Park ...) B: can be anywhere in any direction in the mapping C: to fill at least four ways ( In fact, grid, vertical, horizontal, etc.) D: Metafile classes can be expanded to fill the method Platform: |
Size: 2048 |
Author:zhangbin |
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Description: 在一个划分成网格的操场上，n个士兵散乱地站在网格点上。网格点由整数坐标(x,y)表示。士兵们可以沿网格边上、下、左、右移动一步，但在同一时刻任一网格点上只能有一名士兵。按照军官的命令，士兵们要整齐地列成一个水平队列，即排列成(x,y),(x+1,y),…,(x+n-1,y)。如何选择x和y的值才能使士兵们以最少的总移动步数排成一列。
-In a grid divided into the playground, n soldiers stood scattered points on the grid. Grid point by the integer coordinates (x, y), said. Soldiers along the edge of the grid, down, left and right moving step, but at the same time any point on the grid can only be a soldier. In accordance with the orders of officers, the soldiers should be neatly列成level queue, that is arranged in (x, y), (x+1, y), ..., (x+ n-1, y). How to choose the value of x and y in order to enable the soldiers to a total of at least moves one line up. Platform: |
Size: 1024 |
Author:崔玉迎 |
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Description: 8*8 的国际象棋棋盘上的一只马，恰好走过除起点外的其它 63 个位置各一次，最后回
到起点。这条路线称为一条马的 Hamilton 周游路线。对于给定的 m*n 的国际象棋棋盘，m
和n均为大于5的偶数，且|m-n|≤2，试设计一个分治算法找出一条马的Hamilton周游路线。
编程任务：
对于给定的偶数m，n≥6，且|m-n|≤2，编程计算 m*n 的国际象棋棋盘一条马的Hamilton
周游路线。
Input
输入数据的第一行有2个正整数m和 n，表示给定的国际象棋棋盘
由 m行，每行n个格子组成。
Output
将计算出的马的Hamilton周游路线用下面的2 种表达方式输出。
第 1 种表达方式按照马步的次序给出马的 Hamilton 周游路线。马的每一步用所在的方
格坐标(x，y)来表示。x表示行的坐标，编号为 0，1，…，m-1；y表示列的坐标，编号为 0，
1，…，n-1。起始方格为（0，0）。
第 2 种表达方式在棋盘的方格中标明马到达该方格的步数。（0，0）方格为起跳步，并
标明为第1步。
-8* 8 chess board on a horse, just walk outside than from the other 63 positions each time, and finally back
To the starting point. This line is called Hamilton a horse travel routes. For a given m* n chess board, m
And n are even greater than 5, and | mn | ≤ 2, the partition algorithm to design a test to find a horse of Hamilton travel routes.
Programming tasks:
For a given even number m, n ≥ 6, and | mn | ≤ 2, programming computing m* n chess board a horse Hamilton
Travel routes.
Input
The first line of input data are two positive integers m and n, that a given chessboard
By the m lines of n-grid component.
Output
The calculated travel route horse Hamilton 2 with the following expression of the output.
The first one kind of expression is given in accordance with the order of the horse Horse Hamilton travel routes. Horse with each step of the square where
Grid coordinates (x, y) to represent. the coordinates of x that line, numbered 0,1, ..., m-1 y co Platform: |
Size: 16384 |
Author:zhang |
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Description: 1.生成mfc基于对话框的模块。
2.绘制表格，获取对话框的宽度和高度后画线等分。在onpaint函数内操作的，画棋子也是在其中操作的。
3.根据表格的交界和数组对应，就相当于一个点看作一个元素。同时每个点对应的坐标值是可以算出来的，就是等分原理。比如总长100，坐标就是10，20，。。
4.事件映射数组，有一个双击事件，用来获取鼠标的光标值，这个值和网格点的哪个值最接近就算是哪个点被点击了。
5被点击后数组用1，2标志。onpaint里话棋子，如果为1，2就画两种不同的子。
6，算法，就是看数组中子的情况。比如有五个连续的1，那么就提升我赢了。具体判断可以用一个点一个点的if。
7，是倒推的，如果发现电脑自己有4子，就把盘边的子值设置为2.否则发现自己有3子，且可胜也置2.没中情况都写成函数。具体可见程序。-1. Generate mfc dialog based module.
2. Drawing table, get the width and height of the dialog box after drawing a line decile. Operating within the function in onpaint painting pieces in which the operation is.
3. According to the junction and the array of forms corresponding to the equivalent of a point as an element. While the corresponding coordinates of each point can be counted out, is the attainment theory. Such as total length of 100, the coordinates are 10,20,. .
4. Incident mapping array, a double-click the event, to get the mouse cursor value, the value and the value of the nearest grid point which even the point which was clicked.
5 After the array was clicked with the 1,2 mark. onpaint pieces in it, if it is 1,2 on two different sub-paint.
6, the algorithm is to see an array of neutron situation. For example, five consecutive one, then upgrade and I won. Determine a specific point can be a point of if.
7, is pushed down, and if found to computer they have four c Platform: |
Size: 4722688 |
Author:李龙 |
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Description: 主要功能：
1.动态显示数据。在类中定义了两个大小为2000的double型数值。可显示小于2000组的X,Y数据。动态显示的原理为实时更新数据。只要更新x,y数据，在显示界面就会及时显示。
2.数据捕捉功能。当鼠标靠近数据点时，十字线会绘制一个红色的矩形捕捉框。
3.支持曲线平移，缩放，网格开关，时间显示以及文字说明。平移时，按住右键拖动鼠标；缩放时按住鼠标中键拖动鼠标。
4.支持颜色设置，具体可看弹出菜单。双击鼠标右键，弹出菜单可进行相应的操作。
5.类主要封装了一些设置函数和文件存储读取函数，在具体样例程序中列举了一些函数。具体函数可见点h文件，上面有具体说明。-Main features:
1. Dynamic display data. in the class defines two types of size 2000 double value. Group to show less than 2000 X, Y data. The principle of dynamic display is updated in real time data. As long as the updated x, y data in a timely manner in the display screen will display.
2. Data capture. When the mouse is near the data points, the cross line to draw a red rectangle will capture the box.
3. Support the curve of pan, zoom, grid switch, time display, and text. Pan, hold down the right-drag the mouse zoom, hold down the mouse button drag the mouse.
4. Support the color settings, specifically to see the pop-up menu. Double-click the right mouse button, pop-up menu for the appropriate action.
5. Principal function encapsulates some of the settings and file storage read function, in a specific sample program listed a number of functions. H files visible point specific function, the above specific description. Platform: |
Size: 168960 |
Author:黄雅威 |
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Description: 1） 直线的生成内容：用Bresenham算法画直线要求： 将象素网格表现出来，建立网格坐标系 用橡皮筋的形式输入参数 鼠标移动时，显示鼠标当前位置 显示判别式的计算过程和下一点的选择策略 记录生成点的坐标，建议用表的形式 图形生成过程可以重复进行2）圆弧的生成内容：用Bresenham算法画圆 要求：᠊
-1) line generated content: draw a straight line using the Bresenham algorithm requirements: manifested establish grid coordinate system with a rubber band in the form of input parameters mouse to move the pixel grid, the current mouse position display discriminant calculation process and the selection of the next point strategy record generating point coordinates, it is recommended the table form graphics generation process may be repeated 2) of the arc-generated content: Bresenham algorithm drawcircle requirements: ᠊ Platform: |
Size: 26624 |
Author:假装 |
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Description: This paper presents a proposed model in an
explicit form and its simulation results for
characterizing wideband line-of-sight (LOS)
radio channel in an urban microcellular
environment. The model is fast to compute, easy
to implement, and only those rays that reach the
receiver locations are calculated. Effect of
street-grid, with any number of crossing streets
of different widths, on the propagation
characteristics can be studied. The model
provides details of possible radio paths that join
the basestation with the mobile terminal at any
point in the main street. The characterization of
the channel is presented in three domains. They
are: frequency, temporal and angular domains.
The model can be used to study several
microcellular communication problems. Platform: |
Size: 404480 |
Author:徐清沐 |
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Description: 这个例子展示了如何利用kendoui的grid控件动态改变grid的行的颜色，关键点在于绑定数据源，希望对你们有帮助-This example shows how to use dynamic grid control kendoui to change the grid of the color line, the key point is that the bound data source, and I hope to help you Platform: |
Size: 1024 |
Author:KENNY |
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Description: 实验四 不规则区域的填充算法
实验题目：不规则区域的填充算法
实验目的：验证不规则区域的填充算法
实验内容：利用VC与OpenGL,实现不规则区域的填充算法。
1、必做：实现简单递归的不规则区域填充算法。
2、选做：针对简单递归算法栈空间占用太大的缺点，进行改进，实现基于扫描线的种子填充算法
实验要求：
将坐标系网格在屏幕上画出来，每个像素点占据一个格点，用一个小实心圆圈表示。
用鼠标点击的方式，绘制不规则区域的边界。
种子填充算法，可用4联通或8联通任选一种。
-Filling algorithm experiment four irregular region
Experimental subjects: irregular region filling algorithm
Purpose: Verify that the irregular region filling algorithm
Experiment: Use VC and OpenGL, irregular region filling algorithm realization.
1, will do: simple recursive irregular region filling algorithm.
2, choose to do: take up too much stack space algorithm for simple recursive shortcomings, to improve, to achieve seed filling algorithm based on scan line
Experimental requirements:
The coordinate grid drawn on the screen, each pixel occupies one lattice point, by a small filled circle.
Use the mouse to click on a way to draw irregular border region.
Seed filling algorithm, can be 4 or 8 Unicom Unicom Choose one. Platform: |
Size: 647168 |
Author:赵雯雯 |
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Description: 对这三个面的划分是为了提高网格的密度，对面网格的划分只能在整体的XY平面内进行，所以不能对其他的面进行网格划分来作为体扫掠的依据。这个划分不能得到和右半部分的划分一样的在每条边上10等分，要想得到10等分的划分网格，必须先将角点处的线5等分，然后将角点的面连结，再用体扫掠即可。-These three faces division is to increase the density of the grid, the grid can only be divided across the whole of the XY plane, it can not be meshing on the other side as a body sweep basis. This division can not get the right part of the same and divided in 10 equal portions on each side, in order to get 10 aliquots of mesh, you must first corner point line 5 equal portions, and then the corner points of the surface links , then swept to the body. Platform: |
Size: 1024 |
Author:new |
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Description: Drunkard.java Your Drunkard class. The interface is provided. You will be completing the implementation, and a test driver for it.
RandomWalkViewer.java Your RandomWalkViewer class. You create this file and class.
RandomWalkComponent.java Your RandomWalkComponent class. You create this file and class.
DrunkardTester.java Your unit test program for your Drunkard class. You create this file and class.
ImPoint.java An immutable point class for use in this program. Do not modify or submit this file.
ImPointTester.java Unit test program for ImPoint. Use this to test out ImPoint. Also can be used as an example of what a test program might look like. Do not submit this file.-In this assignment you will write a graphics-based program to do a random walk, sometimes also known as a drunkard s walk. This random walk simulates the wandering of an intoxicated person on a square street grid. The drunkard will start out in the middle of the grid and will randomly pick one of the four compass directions, and take a step in that direction, then another step that new location in a random direction, etc. This will be repeated some number of times determined by the user. The output will display the path of the drunkard as a sequence of line-segments. (This problem is adapted programming problem P6.6 the textbook.)
This assignment will give you practice with creating classes, using loops, using the java library for random number generation, doing console-based IO, and drawing to a graphics window. Also you ll get practice in general program development. Platform: |
Size: 4096 |
Author:Lannister |
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Description: 现有一种格子迷宫，该迷宫由N行N列的格子构成。迷宫中有三种格子，第一种是墙壁，不能通过；第二种格子能正常通行，经过该格子可以到达与它相邻的上/下/左/右格子中；第三种格子除了具有第二种格子的功能外，它还具有“瞬间移动”功能，到达此格子后，可以选择是否瞬间被传送到给定的格子上。在一个由上述三种格子构成的迷宫中，给定起点和终点的位置，求从起点到达终点所需要的最小代价。
下图是一个3*3格子迷宫，该迷宫中2号格子具有的“瞬间移动”功能是能传送至8号格子，7号格子则是能传送至9号格子。
3*3迷宫示例
1 0(8) -1
2 1 2
0(9) 1 2
Input Description
第一行：迷宫的规模N值 起点格子序号start 终点格子序号end
接下来的N*N行：依次按逐行扫描方式输入迷宫中序号为1至N*N的格子的代价值及跳转的格子序号M(如果有的话)。
Output Description
输出从起点到达终点所需要的最小代价。这里需要注意的是，若从指定起点无法到达终点，则输出为INF(请在程序中宏定义为4字节int型变量表示的最大正数值，即为2147483647)。-Present a lattice maze, the labyrinth consists of a grid of N rows and N columns. There are three maze grid, the first one is a wall, you can not pass the second grid normal passage through the lattice can be reached with its adjacent up/down/left/right grid and the third grid in addition to a first two grid functions, it also has " teleport" feature, this lattice arrive, you can choose whether a given moment is transferred to the lattice. In the maze of a grid consisting of the three, given a start and end location, find the minimum cost required the beginning of the end arrives. The figure is a 3* 3 grid maze, the maze has a grid on the 2nd of " teleport" feature is capable of transmitting to the 8th grid, the grid is capable of transmitting the 7th to the 9th grid. Labyrinth Example 3* 3 1 0 (8)-1 2 1 2 0 (9) 1 2 Input Description First line: the value of the starting point of the maze lattice size N number start end end The next grid numbers N* N lines: followe Platform: |
Size: 1024 |
Author:mingren |
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Description: 一个扫雷程序，输入你想要点开的位置的坐标，以及‘a’或‘d’，‘a’相当于左键那个格子就会打开，‘d’相当于右键，那个格子会插上小红旗。输入格式如“3 4 a”然后敲击回车，意思是打开第三行第四个。-A demining program, enter the location you want to point to open the coordinates, and a or d , a is equivalent to the grid will be left open, d is equivalent to the right, the grid will plug a small red flag . Input formats such as 3 4 a then hit Enter, meaning open the third line fourth. Platform: |
Size: 1024 |
Author:徐晃 |
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Description: 设平面上有一个m×n 的网格，将左下角的网格点标记为（0,0）而右上角的网格点标记为（m,n）。某人想从（0,0）出发沿网格线行进到达（m,n），但是在网格点（i,j）处他只能向上行进或者向右行进，向上行进的代价为aij（amj ＝＋∞），向右行进的代价是bij（bin ＝＋∞）。试设计一个动态规划算法，在这个网格中为该旅行者寻找一条代价最小的旅行路线。(There is a m * n mesh on the plane, and the grid point on the lower left is marked (0,0), and the grid point on the upper right is marked as (m, n). Want sb (0,0) starting from marching along the mesh line at (m, n), but in the grid point (I, J) where he can only travel up or to travel on the road to the right, the price for AIJ (AMJ = 2), the road to the right is the price of bij (bin = 2). A dynamic programming algorithm is designed to find the least cost travel route for the traveler in this grid.) Platform: |
Size: 4096 |
Author:16物联网
|
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Description: 运用百度API展示大量的点、线、面数据，每种数据也有不同的展示类型 如直接打点、热力图、网格、聚合等方式展示数据。(Using Baidu API to display a large number of point, line, and surface data, each kind of data also has different display types, such as direct point, thermal diagram, grid, aggregation and other ways to display data.) Platform: |
Size: 5614592 |
Author:黑人的白装 |
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Search - point to point grid line