Question

我需要将三角形打包成一个盒子，尽可能合理，作为3D优化的一部分（我将使用不同纹理的片段填充到一个不同的纹理中，用于深度排序，因此纹理不要每次换新时都要切换

有这样做的算法吗？三角形本身可以制成可叠加的（可转换成直角，有效地使这成为一个盒子填充算法），但我想尽可能避免这种情况，因为它会扭曲底层的纹理艺术。

Answer 1

“紧凑合理” - ＆gt;工作的东西总比没有好。

这些代码片段提供了一种简单的解决方案，可以将形状（也可以是三角形）逐条带填充到矩形中。

public abstract class Shape {
    protected Point offset = new Point();
    public abstract int getHeight();
    public abstract int getWidth();
}

public class Triangle extends Shape {
    // all points are relative to offset (from Shape)
    Point top = new Point(); // top.y is always 0, left.y >= 0 right.y >= 0 
    Point left = new Point(); // left.x < right.x
    Point right = new Point();

    public int getHeight() {
        return left.y >= right.y ? left.y : right.y;
    }

    public int getWidth() {
        int xmin = left.x <= top.x ? left.x : top.x;
        int xmax = right.x >= top.x ? right.x : top.x;
        return xmax - xmin;
    }
}

public class StuffRectangle extends Shape {
    private Point ww = new Point();

    private ArrayList<Shape> maintained = new ArrayList<Shape>();
    private int insx;
    private int insy;
    private int maxy;

    public int getHeight() {
        return ww.y;
    }

    public int getWidth() {
        return ww.x;
    }

    public void clear() {
        insx = 0;
        insy = 0;
        maxy = 0;
        maintained.clear();
    }

    /**
     * Fill the rectangle band by band.
     * 
     * The inserted shapes are removed from the provided shape collection.
     * 
     * @param shapes
     *            the shapes to insert
     * @return the count of inserted shapes.
     */
    public int stuff(Collection<Shape> shapes) {
        int inserted = 0;

        for (;;) {
            int insertedInPass = 0;
            for (Iterator<Shape> i = shapes.iterator(); i.hasNext();) {
                Shape shape = i.next();

                // does the shape fit into current band?
                int sx = shape.getWidth();
                if (insx + sx > getWidth())
                    continue;
                int sy = shape.getHeight();
                if (insy + sy > getHeight())
                    continue;

                // does fit
                ++insertedInPass;

                // remove from shapes
                i.remove();

                // add to maintained and adjust offset
                maintained.add(shape);
                shape.offset.x = insx;
                shape.offset.y = insy;

                insx += sx;
                if (sy > maxy)
                    maxy = sy;

            }
            inserted += insertedInPass;
            if (shapes.isEmpty())
                break;
            // nothing fits current band - try a new band
            if (insertedInPass == 0) {
                // already a new band - does not fit at all
                if (insx == 0)
                    break;

                // start new band
                insx = 0;
                insy += maxy;
                maxy = 0;
                continue;
            }
        }

        return inserted;
    }
}

将任意三角形打包成有限的盒子？

1 个答案: