英文:
How to use the interior area?
问题
关于正交多边形的边缘问题，我的目标是确定一条边是顶边（这意味着多边形的内部位于其下方），底边（这意味着多边形的内部位于其上方），右边（多边形的内部在其左侧）还是左边（多边形的内部在其右侧）。
我有多边形的坐标，并按逆时针方向排序（这是通常推荐的方式）。
我编写了一些用于在Java中计算多边形内部的代码。
public static double polygonArea(double X[], double Y[], int n){// 初始化面积double area = 0.0;// 计算鞋带公式的值int j = n - 1;for (int i = 0; i < n; i++){area += (X[j] + X[i]) * (Y[j] - Y[i]);// j 是 i 的前一个顶点j = i;}// 返回绝对值return Math.abs(area / 2.0);}
... 这在很大程度上是有效的。然而，我绝对不知道如何利用这个来计算或决定一条边（例如，由其顶点和底点定义的垂直边）是左边（内部在右侧）还是右边（内部在左侧）。
也许还有一种更简单的解决方案。然而，如果有人能给我一些建议来解决我的问题，我会非常感谢。
英文:
I have a question regarding the edges of an orthogonal polygon.
My goal is to decide whether an edge is a top edge (this means the interior of the polygon lies below), a bottom edge (this means the interior of the polygon lies above), a right edge (interior of the polygon is left) or a left edge (interior of the polygon is right).
I have the coordinates of the polygon and sorted them counterclockwise (as this is the usual recommended way).
I wrote some code to calculate the interior of the polygon in Java.
[...]
public static double polygonArea(double X[], double Y[], int n){// Initialze area double area = 0.0; `// Calculate value of shoelace formulaint j = n - 1;for (int i = 0; i < n; i++){area += (X[j] + X[i]) * (Y[j] - Y[i]);// j is previous vertex to ij = i;}// Return absolute valuereturn Math.abs(area / 2.0);}
... what is working well. However I have absolutely no clue how I can use this to calculate or decide whether an edge (e.g. a vertical edge defined by its top point and bottom point) is a left edge (interior is right) or a right edge (interior left).
Maybe there is also a way more easier solution. However, I would appreciate if someone could give me some advice to approach my problem.
答案1
得分: 1
如果我正确理解您的问题（图示会有帮助），您可以通过比较 x（水平）或 y（垂直）坐标的顺序，相对于一个简单的（无自交点），正交多边形，获取每条边的位置（顶部、底部、左侧、右侧）。
显然，点的顺序很重要，因此对于逆时针多边形，您可以使用类似以下的方法：
enum EdgeType {TOP, BOTTOM, LEFT, RIGHT, EMPTY}public EdgeType orthoEdgeTypeCCW(double x0, double y0, double x1, double y1){if(x0 == x1) // 垂直{return (y0 < y1) ? EdgeType.RIGHT :(y0 > y1) ? EdgeType.LEFT :EdgeType.EMPTY;}else if(y0 == y1) // 水平{return (x0 < x1) ? EdgeType.BOTTOM :(x0 > x1) ? EdgeType.TOP :EdgeType.EMPTY;}else{throw new IllegalArgumentException("边不正交");}}
（以上为代码部分的翻译。）
英文:
If I've understood your question correctly (a diagram would help) you can get the position (TOP, BOTTOM, LEFT, RIGHT) of each edge relative to a simple (no self-intersections), orthogonal polygon by comparing the ordering of the x (for horizontal) or y (for vertical) coordinates.
Obviously the ordering of the points matters, so for a counter-clockwise polygon you could use something like this:
enum EdgeType {TOP, BOTTOM, LEFT, RIGHT, EMPTY}public EdgeType orthoEdgeTypeCCW(double x0, double y0, double x1, double y1){if(x0 == x1) // vertical{return (y0 < y1) ? EdgeType.RIGHT :(y0 > y1) ? EdgeType.LEFT :EdgeType.EMPTY;}else if(y0 == y1) // horizontal{return (x0 < x1) ? EdgeType.BOTTOM :(x0 > x1) ? EdgeType.TOP :EdgeType.EMPTY;}else{throw new IllegalArgumentException("Edge not orthogonal");}}
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