问题
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 - 9
for 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);
}
}
回答1:
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.
回答2:
You can probably use the Core Java libraries itself:
List<Rectangle2D.Double> rectangles = Arrays.asList(
new Rectangle2D.Double(0d, 0d, 100d, 100d),
new Rectangle2D.Double(100d, 0d, 100d, 100d),
new Rectangle2D.Double(0d, 100d, 100d, 100d),
new Rectangle2D.Double(100d, 100d, 100d, 100d));
Point2D.Double aPoint = new Point2D.Double(30d, 40d);
for (Rectangle2D.Double rectangle:rectangles){
if (rectangle.contains(aPoint)){
System.out.println(rectangle + " has the point " + aPoint);
}
}
回答3:
Working with plane geometry is extremely easy while using JTS. You can try convert the objects you are using to JTS-specific.
来源:https://stackoverflow.com/questions/11817227/check-if-an-array-of-points-is-inside-an-array-of-rectangles