计算两个多边形与cgal的交叉区域

Question

考虑到两个凸多边形的顶点，使用cgal计算交叉区域的最简单方法是什么？

Answer 1

因为您正在使用凸多边形，所以无需担心漏洞。所以我能想到的最简单的代码基本上是构造多边形，调用交集，循环交叉并总计区域::

#include <iostream>
#include <CGAL/Simple_cartesian.h>
#include <CGAL/Polygon_2.h>
#include <CGAL/Polygon_with_holes_2.h>
#include <CGAL/Boolean_set_operations_2.h>
#include <CGAL/Polygon_2_algorithms.h>


typedef CGAL::Simple_cartesian<double> K;
typedef K::Point_2 Point;
typedef CGAL::Polygon_2<K> Polygon_2;
typedef CGAL::Polygon_with_holes_2<K> Polygon_with_holes_2;
using std::cout; using std::endl;


int main(){
  Point points[] = { Point(0,0), Point(1,0), Point(1,1), Point(0,1)};
  Point points2[] = { Point(0.5,0.5), Point(1.5,0.5), Point(1.5,1.5), Point(0.5,1.5)};
  Polygon_2 poly1(points, points+4);
  Polygon_2 poly2(points2, points2+4);
  //CGAL::General_polygon_with_holes_2<K> poly3;
  std::list<Polygon_with_holes_2> polyI;

  CGAL::intersection(poly1, poly2, std::back_inserter(polyI));

  double totalArea = 0;
  typedef std::list<Polygon_with_holes_2>::iterator LIT;
  for(LIT lit = polyI.begin(); lit!=polyI.end(); lit++){
    totalArea+=lit->outer_boundary().area();
  }
  cout << "TotalArea::" << totalArea;

}