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以下代码没有给我预期的结果。我希望将所有点作为多边形的节点，并且多边形包含所有点。如何正确使用 API？此外，如果我将阈值设置为小于 1，程序就会进入某种无限循环。

import com.vividsolutions.jts.geom.Coordinate;
import com.vividsolutions.jts.geom.Geometry;
import com.vividsolutions.jts.geom.GeometryCollection;
import com.vividsolutions.jts.geom.GeometryFactory;
import com.vividsolutions.jts.geom.Point;
import java.util.Set;
import org.geotools.geometry.jts.JTSFactoryFinder;
import org.opensphere.geometry.algorithm.ConcaveHull;


public class ConcaveHullTrial {

    public static void main(String args[]) {

        ConcaveHull cch = null;
        com.vividsolutions.jts.geom.Point point[] = new Point[5];
        Set<GeometryFactory> testing = JTSFactoryFinder.getGeometryFactories();
        for (GeometryFactory f : testing) {
            System.out.println(f);
        }
        GeometryFactory gf = JTSFactoryFinder.getGeometryFactory();
        point[0] = gf.createPoint(new Coordinate(0.0, 0.0));
        point[1] = gf.createPoint(new Coordinate(0.0, 2.0));
        point[2] = gf.createPoint(new Coordinate(2.0, 0.0));
        point[3] = gf.createPoint(new Coordinate(2.0, 2.0));
        point[4] = gf.createPoint(new Coordinate(1.8, 1.5));

        GeometryCollection gc = new GeometryCollection(point, gf);
        double threshold = 4;
        cch = new ConcaveHull(gc, threshold);
        Geometry hull = cch.getConcaveHull();
        for (int i = 0; i < 5; i++) {
            System.out.println(hull.covers(point[i]));
        }
        for (Coordinate c: hull.getCoordinates()) {
            System.out.println(c);
        }
    }
}

结果：

false
true
false
true
false
(0.0, 2.0, NaN)
(2.0, 2.0, NaN)
4

1 回答 1

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这对我来说可以。我使用了 ConcaveHull-0.2 和 geospark-1.1.3 库。这是运行您的代码后的结果：

com.vividsolutions.jts.geom.GeometryFactory@1e81f4dc true true true true true (0.0, 0.0, NaN) (2.0, 0.0, NaN) (2.0, 2.0, NaN) (0.0, 2.0, NaN) (0.0, 0.0, NaN) )

此外，它适用于阈值 < 1.0

于 2019-01-15T02:27:06.310 回答