我想高效地创建一个相交的多边形的顶点编号列表。为了检查是否稍微移动多边形,我可以检查它是相同的交集还是完全不同的交集。大概如果列表中至少有一个顶点号,则它是相同的交点。
因此,我在t = 0处有多边形P和Q相交,这应该给我P的顶点编号[1,2,3]和多边形Q的顶点编号[2],请参见图像进行澄清。在t = 1时,对于P,我将得到[1,2],对于Q,我将得到[2]。
当前,我首先检索交点,然后两个多边形都在顶点之间循环,并使用has_on_boundary函数确定它是否是交点的一部分并存储顶点号。查看代码:
//Standard Includes
#include <iostream>
using std::cout;
using std::endl;
//Include CGAL options
#define CGAL_HEADER_ONLY 1 //Defining to use the CGAL library as header only
#include <CGAL/Exact_predicates_exact_constructions_kernel.h>
#include <CGAL/Boolean_set_operations_2.h>
#include <CGAL/Constrained_Delaunay_triangulation_2.h>
//Typedefs for CGAL Uses.
typedef CGAL::Exact_predicates_exact_constructions_kernel Kernel;
typedef Kernel::Point_2 Point_2;
typedef CGAL::Polygon_2<Kernel> Polygon_2;
typedef CGAL::Polygon_with_holes_2<Kernel> Polygon_with_holes_2;
typedef std::vector<Polygon_with_holes_2> Pwh_list_2;
int main() {
Polygon_2 P;
P.push_back(Point_2(-1, 1));
P.push_back(Point_2(-0.01, -1));
P.push_back(Point_2(0, -1));
P.push_back(Point_2(0.01, -1));
P.push_back(Point_2(1, 1));
Polygon_2 Q;
Q.push_back(Point_2(-1, -1));
Q.push_back(Point_2(1, -1));
Q.push_back(Point_2(0, 1));
Pwh_list_2 polygon_intersections;
CGAL::intersection(P, Q, std::back_inserter(polygon_intersections));
//For ease now only assume one intersection
Polygon_2 intersection = polygon_intersections[0].outer_boundary();
//Get vertex number of P in intersection
int vertex_nr = 0;
std::set<int> overlapping_vertices_P;
cout << "Find intersection vertices of Polygon P" << endl;
for (auto it = P.vertices_begin(); it != P.vertices_end(); ++it, ++vertex_nr) {
if (intersection.has_on_boundary(*it)) {
overlapping_vertices_P.insert(vertex_nr);
cout << vertex_nr << endl;
}
}
//Get vertex number of Q in intersection
vertex_nr = 0;
std::set<int> overlapping_vertices_Q;
cout << "Find intersection vertices of Polygon Q" << endl;
for (auto it = Q.vertices_begin(); it != Q.vertices_end(); ++it, ++vertex_nr) {
if (intersection.has_on_boundary(*it)) {
overlapping_vertices_Q.insert(vertex_nr);
cout << vertex_nr << endl;
}
}
return 0;
}
问题:它可行,但对我来说似乎效率很低。有更好的方法吗?
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