我有顶点列表和区域(其是正方形/矩形)形状的列表。顶点具有x和y坐标,和一个区域具有(X,Y,高度和宽度)。我怎样才能有效地检查哪些顶点位于该地区的每个顶点/地区?
编辑:
这是code我写的做到这一点。
如果(!g.getVertices()的isEmpty()){ 的for(int i = 0; I< g.getVertices()大小();我++){ 顶点v = g.getVertices()得到(一)。 点vertexPoint =新的点(v.getX(),v.getY()); 对于(INT J = 0; J< g.getNumberOfRegions(); J ++){ INT X = g.getRegions()获得(J).getX()。 。INT Y = g.getRegions()获得(J).getY(); INT高度= g.getRegions()获得(J).getHeight()。 INT宽度= g.getRegions()获得(J).getWidth()。 格regionGrid =新格第(j + 1,X,Y,高度,宽度); 矩形regionRectangle =新的Rectangle(X,Y,高度,宽度); 如果(regionRectangle.contains(vertexPoint)){ 的System.out.println(顶点+ V +谎言里地区+ regionGrid.getRegionID()); } } } }编辑2:我用这个生成的地区,但我需要一种方法来从左至右网格中的regionID每个区域分配。例如:
1 - 2 - 34 - 5 - 67 - 8 - 9对于一个3x3的网格。目前,它是以下形式:
1 - 1 - 12 - 2 - 23 - 3 - 3 的for(int i = 0; I< rowValue;我++){ 对于(INT J = 0; J< columnValue; J ++){ 网格R =新的网格(0,20 +我*的大小,20 + J *的大小,尺寸,大小); r.setRegionID第(j + 1); g.addRegion(R); } }解决方案
检查,如果一个顶点是一个正方形内或圆可以在O完成(1)。你可以用库函数或初等数学做到这一点。所以作品的算法可以创建为O(#vertices * #regions)。你可以尝试通过X轴排序顶点和地区,然后通过Y轴进行优化,尽量消除检查肯定返回false。但目前看来,在悲观的情况下,你仍然有O(#vertices * #regions)时间。
I have a list of vertices and a list of regions (which are square/rectangle) shaped. Vertex has x and y coordinates, and a region has (x, y, height and width). How can I efficiently check which vertex lies in which region for every vertex/region?
EDIT:
This is the code I wrote to do this.
if (!g.getVertices().isEmpty()) { for (int i = 0; i < g.getVertices().size(); i++) { Vertex v = g.getVertices().get(i); Point vertexPoint = new Point(v.getX(), v.getY()); for (int j = 0; j < g.getNumberOfRegions(); j++) { int x = g.getRegions().get(j).getX(); int y = g.getRegions().get(j).getY(); int height = g.getRegions().get(j).getHeight(); int width = g.getRegions().get(j).getWidth(); Grid regionGrid = new Grid(j+1, x, y, height, width); Rectangle regionRectangle = new Rectangle(x, y, height, width); if (regionRectangle.contains(vertexPoint)) { System.out.println("Vertex " + v + " lies inside region " + regionGrid.getRegionID()); } } } }EDIT 2: I used this to generate the regions, but I need a way to assign each region in the grid a regionID from left to right. For example:
1 - 2 - 3 4 - 5 - 6 7 - 8 - 9for a 3x3 grid. At the moment it is in the following form:
1 - 1 - 1 2 - 2 - 2 3 - 3 - 3 for (int i = 0; i < rowValue; i++) { for (int j = 0; j < columnValue; j++) { Grid r = new Grid(0, 20 + i * size, 20 + j * size, size, size); r.setRegionID(j + 1); g.addRegion(r); } }解决方案
checking if a vertex is inside a square or a circle can be done in O(1). you can do it with library function or elementary math. so the works algorithm you can create is O(#vertices * #regions). you can try to optimise by sorting the vertices and regions by X-axis and then by Y-axis and try to eliminate checking that for sure return false. but seems that in pessimistic scenario you will still have O(#vertices * #regions) time.
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检查点的数组是矩形的数组里面呢?
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