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我试图确定一个点是否在一个形状内。

我找到了一个可以完成这项工作的算法 http://www.ecse.rpi.edu/Homepages/wrf/Research/Short_Notes/pnpoly.html

我用正方形测试了算法。

正方形的总角 = 4

但它返回给我一个错误的结果。（见下面我的输出结果）在我的代码之后

形状.h

#ifndef __Shape__Shape__
#define __Shape__Shape__

class Shape   {


private:
        int xArray[4];
        int yArray[4];
        int x;
        int y;
public:
        bool inPoly(int x,int y);
        void pointInShape();
};

#endif

形状.cpp

#include "Shape.h"
#include <iostream>

bool Shape::inPoly(int x,int y) {

   xArray[0] = 1;
   xArray[1] = 1;
   xArray[2] = 3;
   xArray[3] = 3;

   yArray[0] = 1;
   yArray[1] = 3;
   yArray[2] = 3;
   yArray[3] = 1;

   int i, j, nvert = 4, c = 0;
   for (i = 0, j = nvert - 1; i < nvert; j = i++) {
       if ( ((yArray[i]>y) != (yArray[j]>y)) &&
           (x < (xArray[j]-xArray[i]) * (y-yArray[i]) / (yArray[j]-yArray[i]) + xArray[i]) )
           c = !c;
   }
   return c;
}


void Shape::pointInShape() {
    std::cout << "results" << std::endl;
    std::cout << inPoly(1,1) << std::endl;
    std::cout << inPoly(1,2) << std::endl;
    std::cout << inPoly(1,3) << std::endl;
    std::cout << inPoly(2,1) << std::endl;
    std::cout << inPoly(2,2) << std::endl;
    std::cout << inPoly(2,3) << std::endl;
    std::cout << inPoly(3,1) << std::endl;
    std::cout << inPoly(3,2) << std::endl;
    std::cout << inPoly(3,3) << std::endl;
}

主文件

#include "Shape.h"
#include <iostream>
int main() {
   Shape shape;
   shape.pointInShape();
}

它返回给我这个输出

   results
   1 <-- (means that 1,1 is is within polygon)
   1 <-- (means that 1,2 is is within polygon)
   0 <-- (means that 1,3 is is not within polygon)
   1 <-- (means that 2,1 is is within polygon)
   1 <-- (means that 2,2 is is within polygon)
   0 <-- (means that 2,3 is is not within polygon)
   0 <-- (means that 3,1 is is not within polygon)
   0 <-- (means that 3,2 is is not within polygon)
   0 <-- (means that 3,3 is is not within polygon)

正确的输出应该只返回 2,2 为真

正确的输出

   results
   0 <-- (means that 1,1 is not within polygon)
   0 <-- (means that 1,2 is not within polygon)
   0 <-- (means that 1,3 is not within polygon)
   0 <-- (means that 2,1 is not within polygon)
   1 <-- (2,2 is is within polygon)
   0 <-- (means that 2,3 is is not within polygon)
   0 <-- (means that 3,1 is is not within polygon)
   0 <-- (means that 3,2 is is not within polygon)
   0 <-- (means that 3,3 is is not within polygon)

有什么建议/建议吗？

4

2 回答 2

3

编写的 PNPOLY 代码实际上是用于浮点数而不是整数。如果您将数组和顶点定义为浮点数，它将为您提供预期的结果。

1.0f 和 2.0f 在浮点数学中不相邻，但 1 和 2 在整数数学中。

应该给你提示的是函数原型

int pnpoly(int nvert, float *vertx, float *verty, float testx, float testy)

此外，这一行：

(testx < (vertx[j]-vertx[i]) * (testy-verty[i]) / (verty[j]-verty[i]) + vertx[i]) )

将在这些除法之后将您的 int 截断为最接近的 int。

如果你真的想使用整数，你必须用更多的空间来定义你的形状，比如 100x100 而不是 2x2。

于 2013-11-06T06:05:03.730 回答
1

根据您的链接和此链接：http: //www.faqs.org/faqs/graphics/algorithms-faq/（请参阅第 2.03 节）您使用的算法仅适用于多边形内部/外部的点. 边界上的点可能返回 1 或 0

如果您想知道某个点何时恰好在边界上，则需要另一个程序。这只是 PNPOLY 缺乏的众多功能之一；它也不能预测明天的天气。您可以自由扩展 PNPOLY 的源代码。

于 2013-11-06T05:53:48.103 回答