检测多个多边形交叉口(java)

Question

背景:本项目涉及GIS环境下的区域制图。我需要找到一组多边形和另一组多边形的交集。我通过将与地图兼容的多边形转换为Polygon2d (http://geom-java.sourceforge.net/api/math/geom2d/polygon/package-summary.html)来实现这一点，并使用交集方法为交叉多边形查找顶点列表。

问题:这是可行的，除非在同一两个多边形之间有多个重叠区域。顶点列表是正确的，但我需要将列表划分为每个单独的多边形。

第一个图像应该是什么样子，第二个是实际生成的图像：

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(忽略顶部的区域。这是一个不同的错误的结果)

如何检测和纠正这种情况？

Answer 1

我想出来了。我取交多边形(蓝色)的顶点列表，迭代每对点，将两个原始多边形中的每个线段转换为LineSegment2D对象，并使用.contains方法检查每对点是否位于其中一个线段中。如果每一对点不包含在1行内，那么就会有一个错误。

这里是方法，但请记住，其中涉及到几个专有类。

public static boolean noErrors(SurfacePolygonX p1, SurfacePolygonX p2, List<LatLon> list) {
    boolean allPointsInSamePolygon = true;

    //for each latlon jl in locations, cast jl and jl2 as points    
    for (int j = 0; j < list.size(); j++) {
        LatLon jl = list.get(j);
        LatLon jl2 = list.get((j == list.size() - 1) ? 0 : j + 1);
        Point2D pt = new Point2D(jl.longitude.degrees, jl.latitude.degrees);
        Point2D pt2 = new Point2D(jl2.longitude.degrees, jl2.latitude.degrees);

        List<LatLon> corners = p1.getCorners();

        boolean bothPointsInSameSegment = false;

        //for each latlon k selectedShape cast k and k+1 as 2lineseg
        for (int k = 0; k < corners.size(); k++) {
            LatLon kl = corners.get(k);
            LatLon kl2 = corners.get((k == corners.size() - 1) ? 0 : k + 1);

            LineSegment2D segment = new LineSegment2D(kl.longitude.degrees, kl.latitude.degrees, kl2.longitude.degrees, kl2.latitude.degrees);

            boolean segContainsP1 = segment.contains(pt);
            boolean segContainsP2 = segment.contains(pt2);

            System.out.println("selectedShape: segment "+k+" contains p"+j+":("+segContainsP1+") and p"+(j+1)+":("+segContainsP2+")");

            //check if each line contains the points.
            if (segContainsP1 && segContainsP2)
            {
                bothPointsInSameSegment = true;
            }

        }

        corners = p2.getCorners();
        //for each latlon k tempShape cast k and k+1 as 2lineseg
        for (int k = 0; k < corners.size(); k++) {
            LatLon kl = corners.get(k);
            LatLon kl2 = corners.get((k == corners.size() - 1) ? 0 : k + 1);

            LineSegment2D segment = new LineSegment2D(kl.longitude.degrees, kl.latitude.degrees, kl2.longitude.degrees, kl2.latitude.degrees);


            boolean segContainsP1 = segment.contains(pt);
            boolean segContainsP2 = segment.contains(pt2);

            System.out.println("intersectingShape: segment "+k+" contains p"+j+":("+segContainsP1+") and p"+(j+1)+":("+segContainsP2+")");

            //check if each line contains the points.
            if (segContainsP1 && segContainsP2)
            {
                bothPointsInSameSegment = true;
            }

        }

        if (!bothPointsInSameSegment) allPointsInSamePolygon = false;

    }

    //if both points are not in the same line, then theres a conflict
    return allPointsInSamePolygon;
}

Answer 2

您可以使用JTS。

1. 使用LinearRing创建多边形
2. 使用交叉点方法

简单的代码示例：

// build polygon p1
LinearRing p1 = new GeometryFactory().createLinearRing(new Coordinate[]{new Coordinate(0,0), new Coordinate(0,10), new Coordinate(10,10), new Coordinate(10,0), new Coordinate(0,0)});
// build polygon p2
LinearRing p2 = new GeometryFactory().createLinearRing(new Coordinate[]{new Coordinate(5,5), new Coordinate(15,5), new Coordinate(15,15), new Coordinate(5,15), new Coordinate(5,5)});
// calculate intersecting points
Geometry intersectingPoints = p1.intersection(p2);
// print result
for(Coordinate c : intersectingPoints.getCoordinates()){
    System.out.println(c.toString());
}

输出(与预期的一样)：

(5.0, 10.0, NaN)
(10.0, 5.0, NaN)