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Uniform grid collision detection between circles i

2019-08-04 09:03发布

问题:

I am working on a 2d arcade game where I have 5 types of circles with different sizes: The ship, the missiles, and 3 types of monsters.

This is what it looks like:

Currently I'm using brute force collision detection where I check every missile vs. every monster without taking probability of collision into account. Sadly, this makes the process REALLY slow.

This here is my Grid class, but it's incomplete. I'd greatly appreciate your help.

    public class Grid {

    int rows;
    int cols;
    double squareSize;
    private ArrayList<Circle>[][] grid;

    public Grid(int sceneWidth, int sceneHeight, int squareSize) {
        this.squareSize = squareSize;
// Calculate how many rows and cols for the grid.
        rows = (sceneHeight + squareSize) / squareSize;
        cols = (sceneWidth + squareSize) / squareSize;
// Create grid
        this.grid = (ArrayList[][]) new ArrayList[cols][rows]; //Generic array creation error workaround
    }

The addObject method inside the Grid class.
    public void addObject(Circle entity) {
// Adds entity to every cell that it's overlapping with.
        double topLeftX = Math.max(0, entity.getLayoutX() / squareSize);
        double topLeftY = Math.max(0, entity.getLayoutY() / squareSize);
        double bottomRightX = Math.min(cols - 1, entity.getLayoutX() + entity.getRadius() - 1) / squareSize;
        double bottomRightY = Math.min(rows - 1, entity.getLayoutY() + entity.getRadius() - 1) / squareSize;

        for (double x = topLeftX; x < bottomRightX; x++) {
            for (double y = topLeftY; y < bottomRightY; y++) {
                grid[(int) x][(int) y].add(entity); //Cast types to int to prevent loosy conversion type error.
            }
        }
    }

But that's where I am at a complete loss. I'm not even sure the source code I provided is correct. Please let me know how to make the grid based collision work. I've read basically every tutorial I could get my hands on but without much effect. Thanks.