polygon

What is the quickest way to find the shortest cartesian distance between two polygons

不羁的心 提交于 2019-12-02 18:14:10
I have 1 red polygon say and 50 randomly placed blue polygons - they are situated in geographical 2D space . What is the quickest/speediest algorithim to find the the shortest distance between a red polygon and its nearest blue polygon? Bear in mind that it is not a simple case of taking the points that make up the vertices of the polygon as values to test for distance as they may not necessarily be the closest points. So in the end - the answer should give back the closest blue polygon to the singular red one. This is harder than it sounds! I doubt there is better solution than calculating

highlight a polygon while on click using kml

余生长醉 提交于 2019-12-02 15:11:56
问题 I am using kml file to draw polygons in google map. I am setting click events for those polygon which is working. Now I want the clicked polygon to be highlighted. I had tried the setOption() method, but it didn't work. Is it possible to change the color of polygon on clicking it? 回答1: If you are using KmlLayer (a guess), you can't change the properties of the Polygons. If you use a third party KML parser like geoxml3 or geoxml-v3 to render the polygons as native Google Maps API v3 objects

Box2d custom polygon and sprites mis-matching

僤鯓⒐⒋嵵緔 提交于 2019-12-02 13:22:59
问题 I am using the Physics Editor for creating the Polygon in Box2d. It generates the Polygon and works fine in the non-retina display but doesn't work in the retina display..... I have attached the screen shot for both of the displays.Now when comes to retina display the polygon is not set over the car here's the image for that Here's my code which I am using in the Project CCSprite *car = [CCSprite spriteWithFile:@"opp_car.png"]; [car setPosition:ccp(wSize.width/2+50,wSize.height/2-120)]; [self

How to define an n-sided polygon function in haskell

纵饮孤独 提交于 2019-12-02 13:19:40
问题 I need to define a function that moves the "pen", which starts at (0,0) in the bottom left, around to draw the correct polygon depending on the input, with every side of the shape having length of 1. I need to do it without importing any functions that already exist. I am using OpenGL to draw the shape. Command is defined as: data Command = Fd Double | Bk Double | Lt Double | Rt Double | Go Command | Command :> Command I have done a triangle which works but now want to be able to just say how

Rotating a polygon in Java

六月ゝ 毕业季﹏ 提交于 2019-12-02 09:49:38
The program I am writing draws multiple stars on the screen and gives them random directions and speeds. The stars will bounce off the edges of the panel and stay inside. I need to have the stars rotating as they are moving. I have tried so many things and I cannot figure it out. Below is the code I use to draw the stars and move them. Additional Information: -the stars are in a collection called "stars" -there are two classes that I wrote, "Star" and "MyJPanel" Set the points: (in Star class) for (double current = 0; current < nPoints; current += 1) { i = (int)current; int[] X = new int

How to Apply transformation to Polygon object in Java

こ雲淡風輕ζ 提交于 2019-12-02 08:41:00
i have made a transform and rendered a Polygon object with it(mesh is of type Polygon): at.setToTranslation(gameObject.position.x, gameObject.position.y); at.rotate(Math.toRadians(rotation)); at.scale(scale, scale); g2d.setTransform(at); g2d.fillPolygon(mesh); now i want to return the exact mesh i rendered so that i can do collision checks on it. only problem is that if i return mesh it returns the un-transformed mesh. so i tried setting the transform to the Polygon object (mesh) like so: mesh = (Polygon)at.createTransformedShape(mesh); but unfortunately at.createTransformedShape() returns a

Sort point list into polygon

妖精的绣舞 提交于 2019-12-02 08:27:18
问题 I have a set of points. This set of points do define a (non convex) polygon but its not ordered. Since it's not ordered I cannot just draw from point to point to draw its border. How can I sort it in a way I can walk through this point list and draw a polygon? My first idea was to use a convex hull but my polygons are, most of the time, concave. 回答1: I don't think there's a well-defined solution to this. Consider five points like this: . . . . . What polygon would be correct here? 回答2: You

Sort point list into polygon

白昼怎懂夜的黑 提交于 2019-12-02 08:01:01
I have a set of points. This set of points do define a (non convex) polygon but its not ordered. Since it's not ordered I cannot just draw from point to point to draw its border. How can I sort it in a way I can walk through this point list and draw a polygon? My first idea was to use a convex hull but my polygons are, most of the time, concave. I don't think there's a well-defined solution to this. Consider five points like this: . . . . . What polygon would be correct here? You have to order the points so you walk around the polygon with the interior on your left (or right) as you move from

Generating a filled polygon inside a numpy array

假装没事ソ 提交于 2019-12-02 07:29:05
问题 I'm looking for a way to 'draw' a filled polygon into a numpy array based upon a set of polygon vertices. I'd prefer to use as few external libraries as possible. For example: I have a 20x20 numpy array and I'd like the region bounded by points (3,12), (8,18), (13,14), (11,6) and (4,6) to be filled with 1.0 while the rest of the array contains 0.0 回答1: The following solution requires only numpy. It works for vertices (defined in clockwise order in [Row, Column] coordinate system) for convex

Qt: Extract intensity values clipped by a region of interest polygon

做~自己de王妃 提交于 2019-12-02 07:25:43
Based on a grayscale image and an ordered closed polygon (may be concave), I want to get all grayscale values that lie inside a region of interest polygon (as likewise described in SciPy Create 2D Polygon Mask ). What is the most performant realisation of that in Qt 4.8? Endpoint should be some kind of QList<double> . Thanks for your advices. In addition, is it possible to compute a floating point mask (e.g. 0 for outside the polygon, 0.3 for 30% of the pixel area is within the polygon, 1 for completely inside the polygon)? However, that's just an extra, endpoint would be QPair<double, double>