Поиск, если Path2D самопересекается

Question

Мне нужно найти, пересекает ли Path2D себя. Сейчас я делаю это, просто извлекая массив строк из пути и обнаруживая, пересекаются ли они. Но он имеет сложность O (n ^ 2) и поэтому очень медленный. Есть ли более быстрый способ сделать это?

Answer 1

Вот моя Java реализация этого алгоритма:

import java.awt.Point;
import java.awt.geom.Line2D;
import java.awt.geom.PathIterator;
import java.util.*;

/**
 * Path2D helper functions.
 * <p/>
 * @author Gili Tzabari
 */
public class Path2Ds
{
    /**
     * Indicates if a Path2D intersects itself.
     * <p/>
     * @return true if a Path2D intersects itself
     */
    public static boolean isSelfIntersecting(PathIterator path)
    {
        SortedSet<Line2D> lines = getLines(path);
        if (lines.size() <= 1)
            return false;

        Set<Line2D> candidates = new HashSet<Line2D>();
        for (Line2D line: lines)
        {
            if (Double.compare(line.getP1().distance(line.getP2()), 0) <= 0)
            {
                // Lines of length 0 do not cause self-intersection
                continue;
            }
            for (Iterator<Line2D> i = candidates.iterator(); i.hasNext();)
            {
                Line2D candidate = i.next();

                // Logic borrowed from Line2D.intersectsLine()
                int lineRelativeToCandidate1 = Line2D.relativeCCW(line.getX1(), line.getY1(), line.
                    getX2(),
                    line.getY2(), candidate.getX1(), candidate.getY1());
                int lineRelativeToCandidate2 = Line2D.relativeCCW(line.getX1(), line.getY1(), line.
                    getX2(),
                    line.getY2(), candidate.getX2(), candidate.getY2());
                int candidateRelativeToLine1 = Line2D.relativeCCW(candidate.getX1(),
                    candidate.getY1(),
                    candidate.getX2(), candidate.getY2(), line.getX1(), line.getY1());
                int candidateRelativeToLine2 = Line2D.relativeCCW(candidate.getX1(),
                    candidate.getY1(),
                    candidate.getX2(), candidate.getY2(), line.getX2(), line.getY2());
                boolean intersection = (lineRelativeToCandidate1 * lineRelativeToCandidate2 <= 0)
                    && (candidateRelativeToLine1 * candidateRelativeToLine2 <= 0);
                if (intersection)
                {
                    // Lines may share a point, so long as they extend in different directions
                    if (lineRelativeToCandidate1 == 0 && lineRelativeToCandidate2 != 0)
                    {
                        // candidate.P1 shares a point with line
                        if (candidateRelativeToLine1 == 0 && candidateRelativeToLine2 != 0)
                        {
                            // line.P1 == candidate.P1
                            continue;
                        }
                        if (candidateRelativeToLine1 != 0 && candidateRelativeToLine2 == 0)
                        {
                            // line.P2 == candidate.P1
                            continue;
                        }
                        // else candidate.P1 intersects line
                    }
                    else if (lineRelativeToCandidate1 != 0 && lineRelativeToCandidate2 == 0)
                    {
                        // candidate.P2 shares a point with line
                        if (candidateRelativeToLine1 == 0 && candidateRelativeToLine2 != 0)
                        {
                            // line.P1 == candidate.P2
                            continue;
                        }
                        if (candidateRelativeToLine1 != 0 && candidateRelativeToLine2 == 0)
                        {
                            // line.P2 == candidate.P2
                            continue;
                        }
                        // else candidate.P2 intersects line
                    }
                    else
                    {
                        // line and candidate overlap
                    }
                    return true;
                }
                if (candidate.getX2() < line.getX1())
                    i.remove();
            }
            candidates.add(line);
        }
        return false;
    }


    /**
     * Returns all lines in a path. The lines are constructed such that the starting point is found
     * on the left (or same x-coordinate) of the ending point.
     * <p/>
     * @param path the path
     * @return the lines, sorted in ascending order of the x-coordinate of the starting point and
     * ending point, respectively
     */
    private static SortedSet<Line2D> getLines(PathIterator path)
    {
        double[] coords = new double[6];
        SortedSet<Line2D> result = new TreeSet<Line2D>(new Comparator<Line2D>()
        {
            @Override
            public int compare(Line2D o1, Line2D o2)
            {
                int result = Double.compare(o1.getX1(), o2.getX1());
                if (result == 0)
                {
                    // Ensure we are consistent with equals()
                    return Double.compare(o1.getX2(), o2.getX2());
                }
                return result;
            }
        });
        if (path.isDone())
            return result;
        int type = path.currentSegment(coords);
        assert (type == PathIterator.SEG_MOVETO): type;
        Point.Double startPoint = new Point.Double(coords[0], coords[1]);
        Point.Double openPoint = startPoint;
        path.next();

        while (!path.isDone())
        {
            type = path.currentSegment(coords);
            assert (type != PathIterator.SEG_CUBICTO && type != PathIterator.SEG_QUADTO): type;
            switch (type)
            {
                case PathIterator.SEG_MOVETO:
                {
                    openPoint = startPoint;
                    break;
                }
                case PathIterator.SEG_CLOSE:
                {
                    coords[0] = openPoint.x;
                    coords[1] = openPoint.y;
                    break;
                }
            }
            Point.Double endPoint = new Point.Double(coords[0], coords[1]);
            if (Double.compare(startPoint.getX(), endPoint.getX()) < 0)
                result.add(new Line2D.Double(startPoint, endPoint));
            else
                result.add(new Line2D.Double(endPoint, startPoint));
            path.next();
            startPoint = endPoint;
        }
        return result;
    }
}

Answer 2

Вы можете сделать это быстрее, используя алгоритм строчной развертки: http://en.wikipedia.org/wiki/Sweep_line_algorithm

Псевдокод:

Each line has a start point and an end point. Say that `start_x` <= `end_x` for all the lines.
Create an empty bucket of lines.
Sort all the points by their x coordinates, and then iterate through the sorted list.
If the current point is a start point, test its line against all the lines in the bucket, and then add its line to the 
bucket.
if the current point is an end point, remove its line from the bucket.

Худший случай все еще O(N^2), но средний случай O(NlogN)