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我正在使用 Clipper 并想确定两个（多）多边形是否相交。

我的期望是图书馆会有一种很好的、​​抽象的方式来提出这个问题，但似乎没有。

我认为该Area()方法可能有用，但它仅适用于Path该Execute()方法返回Paths。

我已经构建了以下 M(几乎)WE 来演示该问题：

#include <iostream>
#include "clipper.hpp"
using namespace ClipperLib;

Paths MakeBox(int xmin, int xmax, int ymin, int ymax){
  Paths temp(1);
  temp[0] << IntPoint(xmin,ymin) << IntPoint(xmax,ymin) << IntPoint(xmax,ymax) << IntPoint(xmin,ymax);
  return temp;
}

bool Intersects(const Paths &subj, const Paths &clip){
  ClipperLib::Clipper c;

  c.AddPaths(subj, ClipperLib::ptSubject, true);
  c.AddPaths(clip, ClipperLib::ptClip,    true);

  ClipperLib::Paths solution;
  c.Execute(ClipperLib::ctIntersection, solution, ClipperLib::pftNonZero, ClipperLib::pftNonZero);

  return Area(solution);
}

int main(){
  Paths subj  = MakeBox(0,10,0,10);
  Paths clip1 = MakeBox(1,2,1,2);
  Paths clip2 = MakeBox(15,20,15,20);

  Intersects(subj,clip1);
  Intersects(subj,clip2);
}
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1 回答 1

4

似乎最简单的方法是计算方法Paths返回的对象中的路径数Execute()。Paths是一个简单向量，因此，如果它有size()==0，则不存在交集。

#include <iostream>
#include "clipper.hpp"
using namespace ClipperLib;

Paths MakeBox(int xmin, int xmax, int ymin, int ymax){
  Paths temp(1);
  temp[0] << IntPoint(xmin,ymin) << IntPoint(xmax,ymin) << IntPoint(xmax,ymax) << IntPoint(xmin,ymax);
  return temp;
}

bool Intersects(const Paths &subj, const Paths &clip){
  ClipperLib::Clipper c;

  c.AddPaths(subj, ClipperLib::ptSubject, true);
  c.AddPaths(clip, ClipperLib::ptClip,    true);

  ClipperLib::Paths solution;
  c.Execute(ClipperLib::ctIntersection, solution, ClipperLib::pftNonZero, ClipperLib::pftNonZero);

  return solution.size()!=0;
}

int main(){
  Paths subj  = MakeBox(0,10,0,10);
  Paths clip1 = MakeBox(1,2,1,2);
  Paths clip2 = MakeBox(15,20,15,20);

  Intersects(subj,clip1);
  Intersects(subj,clip2);
}
于 2016-07-14T19:35:32.870 回答