WebGIS中一些功能算法实例
2018-07-20 来源:open-open
1、如何判断平面内两条线是否相交并返回交点?
/// <summary> /// 判断平面内两条线是否相交并返回交点 /// </summary> /// <param name="a">线段1起点坐标</param> /// <param name="b">线段1终点坐标</param> /// <param name="c">线段2起点坐标</param> /// <param name="d">线段2终点坐标</param> /// <param name="intersection">相交点坐标</param> /// <returns>是否相交 0:两线平行 -1:不平行且未相交 1:两线相交</returns> public static int GetIntersection(ESRI.ArcGIS.Client.Geometry.MapPoint a, ESRI.ArcGIS.Client.Geometry.MapPoint b, ESRI.ArcGIS.Client.Geometry.MapPoint c, ESRI.ArcGIS.Client.Geometry.MapPoint d, ref ESRI.ArcGIS.Client.Geometry.MapPoint intersection) { a = new MapPoint(Math.Round(a.X, 3), Math.Round(a.Y, 3)); b = new MapPoint(Math.Round(b.X, 3), Math.Round(b.Y, 3)); c = new MapPoint(Math.Round(c.X, 3), Math.Round(c.Y, 3)); d = new MapPoint(Math.Round(d.X, 3), Math.Round(d.Y, 3)); //判断异常 if (Math.Abs(b.X - a.Y) + Math.Abs(b.X - a.X) + Math.Abs(d.Y - c.Y) + Math.Abs(d.X - c.X) == 0) { /* if (c.X - a.X == 0) Debug.Print("ABCD是同一个点!"); else Debug.Print("AB是一个点,CD是一个点,且AC不同!"); */ return 0; } if (Math.Abs(b.Y - a.Y) + Math.Abs(b.X - a.X) == 0) { /* if ((a.X - d.X) * (c.Y - d.Y) - (a.Y - d.Y) * (c.X - d.X) == 0) Debug.Print("A、B是一个点,且在CD线段上!"); else Debug.Print("A、B是一个点,且不在CD线段上!"); */ return 0; } if (Math.Abs(d.Y - c.Y) + Math.Abs(d.X - c.X) == 0) { /* if ((d.X - b.X) * (a.Y - b.Y) - (d.Y - b.Y) * (a.X - b.X) == 0) Debug.Print("C、D是一个点,且在AB线段上!"); else Debug.Print("C、D是一个点,且不在AB线段上!"); */ return 0; } if ((b.Y - a.Y) * (c.X - d.X) - (b.X - a.X) * (c.Y - d.Y) == 0) { //Debug.Print("线段平行,无交点!"); return 0; } intersection.X = ((b.X - a.X) * (c.X - d.X) * (c.Y - a.Y) - c.X * (b.X - a.X) * (c.Y - d.Y) + a.X * (b.Y - a.Y) * (c.X - d.X)) / ((b.Y - a.Y) * (c.X - d.X) - (b.X - a.X) * (c.Y - d.Y)); intersection.Y = ((b.Y - a.Y) * (c.Y - d.Y) * (c.X - a.X) - c.Y * (b.Y - a.Y) * (c.X - d.X) + a.Y * (b.X - a.X) * (c.Y - d.Y)) / ((b.X - a.X) * (c.Y - d.Y) - (b.Y - a.Y) * (c.X - d.X)); if ((intersection.X - a.X) * (intersection.X - b.X) <= 0 && (intersection.X - c.X) * (intersection.X - d.X) <= 0 && (intersection.Y - a.Y) * (intersection.Y - b.Y) <= 0 && (intersection.Y - c.Y) * (intersection.Y - d.Y) <= 0) { // Debug.Print("线段相交于点(" + intersection.X + "," + intersection.Y + ")!"); return 1; //'相交 } else { //Debug.Print("线段相交于虚交点(" + intersection.X + "," + intersection.Y + ")!"); return -1; //'相交但不在线段上 } }
标签:
版权申明:本站文章部分自网络,如有侵权,请联系:west999com@outlook.com
特别注意:本站所有转载文章言论不代表本站观点!
本站所提供的图片等素材,版权归原作者所有,如需使用,请与原作者联系。
下一篇:Android连接网络的代码
最新资讯
热门推荐