Search - point to point grid line

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[Other] mymapp DL : 0: 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
Date : 2025-12-29 Size : 40kb User : 张建
[Other] zoumigongsheji DL : 0: 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).
Date : 2025-12-29 Size : 1kb User : 肖影
[Other] FPoint DL : 0: 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
Date : 2025-12-29 Size : 4.5mb User : 李龙
[Other] dynamic programming DL : 0: 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.)
Date : 2025-12-29 Size : 4kb User : 16物联网
[Other] web-demo-v3-master DL : 0: 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.)
Date : 2025-12-29 Size : 5.35mb User : 黑人的白装