c++ - 为什么我没有得到这些多边形的任何交集?
问题描述
我学习了如何将 boost::polygon 库与自定义多边形一起使用。我举了一个使用自定义多边形的例子,并试图得到这些的交叉点。我没有任何交集,也不明白为什么。此外,您可以看到这个构造函数
CPoint(boost::polygon::point_data<int> pd)
没有它我无法编译我的程序。有人可以解释一下,为什么我需要它?我做错了什么?
/*
Copyright 2008 Intel Corporation
Use, modification and distribution are subject to the Boost Software License,
Version 1.0. (See accompanying file LICENSE_1_0.txt or copy at
http://www.boost.org/LICENSE_1_0.txt).
*/
#include <boost/polygon/polygon.hpp>
#include <cassert>
#include <list>
namespace gtl = boost::polygon;
using namespace boost::polygon::operators;
template <typename Polygon>
void test_polygon()
{
typedef typename gtl::polygon_traits<Polygon>::point_type Point;
Point pts[] = {gtl::construct<Point>(6, 0),
gtl::construct<Point>(0, 6),
gtl::construct<Point>(-6, 0),
gtl::construct<Point>(0, -6),
};
Polygon poly, poly2;
gtl::set_points(poly, pts, pts+5);
Point pts2[] = {gtl::construct<Point>(4, 0),
gtl::construct<Point>(4, 4),
gtl::construct<Point>(0, 4),
gtl::construct<Point>(0, 0),
};
gtl::set_points(poly2, pts2, pts2+5);
std::vector<Polygon> res;
//boost::polygon::polygon_set_data<int> res;
//res += poly;
//assign(res, poly);
res.push_back(poly);
res &= poly2;
std::cout << "size = " << res.size() << std::endl;
assert(!res.empty());
//for(auto it = res[0].begin(); it != res[0].end(); ++it)
// std::cout << "Point(" << it->x << ", " << it->y << ")" << std::endl;
assert(gtl::area(poly) == 100.0f);
}
struct CPoint {
CPoint()=default;
CPoint(boost::polygon::point_data<int> pd) /*---> WHY? What for ? I implemented constructor with traits*/
{
x = pd.x();
y = pd.y();
}
int x;
int y;
};
namespace boost { namespace polygon {
template <>
struct geometry_concept<CPoint> { typedef point_concept type; };
template <>
struct point_traits<CPoint> {
typedef int coordinate_type;
static inline coordinate_type get(const CPoint& point,
orientation_2d orient) {
if(orient == HORIZONTAL)
return point.x;
return point.y;
}
};
template <>
struct point_mutable_traits<CPoint> {
typedef int coordinate_type;
static inline void set(CPoint& point, orientation_2d orient, int value) {
if(orient == HORIZONTAL)
point.x = value;
else
point.y = value;
}
static inline CPoint construct(int x_value, int y_value) {
CPoint retval;
retval.x = x_value;
retval.y = y_value;
return retval;
}
};
}
}
typedef std::list<CPoint> CPolygon;
namespace boost {
namespace polygon {
template <>
struct geometry_concept<CPolygon>{ typedef polygon_concept type; };
template <>
struct polygon_traits<CPolygon> {
typedef int coordinate_type;
typedef CPolygon::const_iterator iterator_type;
typedef CPoint point_type;
static inline iterator_type begin_points(const CPolygon& t) {
return t.begin();
}
static inline iterator_type end_points(const CPolygon& t) {
return t.end();
}
static inline std::size_t size(const CPolygon& t) {
return t.size();
}
static inline winding_direction winding(const CPolygon& t) {
return clockwise_winding;
}
};
template <>
struct polygon_mutable_traits<CPolygon> {
template <typename iT>
static inline CPolygon& set_points(CPolygon& t,
iT input_begin, iT input_end) {
t.clear();
t.insert(t.end(), input_begin, input_end);
return t;
}
};
}
}
int main() {
test_polygon<CPolygon>();
return 0;
}
我在交叉点的结果向量中得到 size = 0 。
解决方案
gtl::set_points(poly, pts, pts + 5);
这超出了范围,因为数组只有 4 个元素。通过现代化避免整个错误类别:
std::array const pts1{
gtl::construct<Point>(6, 0),
gtl::construct<Point>(0, 6),
gtl::construct<Point>(-6, 0),
gtl::construct<Point>(0, -6),
};
std::array const pts2{
gtl::construct<Point>(4, 0),
gtl::construct<Point>(4, 4),
gtl::construct<Point>(0, 4),
gtl::construct<Point>(0, 0),
};
进而
Polygon poly1;
gtl::set_points(poly1, pts1.begin(), pts1.end());
Polygon poly2;
gtl::set_points(poly2, pts2.begin(), pts2.end());
接下来你有:
static inline winding_direction winding(const CPolygon& t)
{
return clockwise_winding;
}
但是您的示例数据不符合该绕组。修复您的输入数据:
std::array const pts1{
gtl::construct<Point>(0, -6),
gtl::construct<Point>(-6, 0),
gtl::construct<Point>(0, 6),
gtl::construct<Point>(6, 0),
};
std::array const pts2{
gtl::construct<Point>(0, 0),
gtl::construct<Point>(0, 4),
gtl::construct<Point>(4, 4),
gtl::construct<Point>(4, 0),
};
最后,Boost Polygon 中算子的结果是一个多边形集合视图。您可以让图书馆指定:
std::vector<Polygon> res;
gtl::assign(res, poly1 & poly2);
或者
std::vector<Polygon> res{poly1};
res &= poly2;
两者实际上是等效的,尽管第二个可能更有效?
fmt::print("poly1 {}, area {}\n", poly1, gtl::area(poly1));
fmt::print("poly2 {}, area {}\n", poly2, gtl::area(poly2));
fmt::print("res {}, area {}\n", res, gtl::area(res));
印刷
poly1 [(0, -6), (-6, 0), (0, 6), (6, 0)], area 72
poly2 [(0, 0), (0, 4), (4, 4), (4, 0)], area 16
res [[(4, 2), (2, 4), (0, 4), (0, 0), (4, 0), (4, 2)]], area 0
固定区域
我现在唯一看到的是结果集上的缠绕仍然是错误的。
毫无疑问,这就是area(res)
失败的原因。我们可以通过将绕组更改为逆时针来显示(当然还有示例数据):
poly1 {(6, 0), (0, 6), (-6, 0), (0, -6)}, area 72
poly2 {(4, 0), (4, 4), (0, 4), (0, 0)}, area 16
res {{(4, 2), (2, 4), (0, 4), (0, 0), (4, 0), (4, 2)}}, area 14
也许您毕竟想要逆时针(考虑到测试数据),或者可能需要另一个特征来告诉库vector<CPolygon>
作为多边形集的绕组。
修复构造函数问题
当构造函数被移除时,在编译器错误中向上调用链,我们看到
polygon_set_mutable_traits<polygon_set_type_1>::set(lvalue, ps.begin(), ps.end());
这似乎证实了上述预感:您缺少多边形集的特征。但是,添加一些
template <> struct polygon_set_traits<std::vector<CPolygon> > {
using C = std::vector<CPolygon>;
using iterator_type = C::const_iterator;
using coordinate_type = polygon_traits<CPolygon>::coordinate_type;
static iterator_type begin(C const& c) { return c.begin(); }
static iterator_type end(C const& c) { return c.end(); }
static bool clean(C const& c) { return c.empty(); }
};
template <> struct polygon_set_mutable_traits<std::vector<CPolygon> > {
using C = std::vector<CPolygon>;
using iterator_type = C::iterator;
using coordinate_type = polygon_set_traits<C>::coordinate_type;
static iterator_type begin(C& c) { return c.begin(); }
static iterator_type end(C& c) { return c.end(); }
template <typename II>
static inline void set(C& c, II input_begin, II input_end)
{
c.clear();
size_t num_ele = std::distance(input_begin, input_end);
c.reserve(num_ele);
polygon_set_data<coordinate_type> ps;
ps.reserve(num_ele);
ps.insert(input_begin, input_end);
ps.get(c);
}
};
结果相同,因为ps.get
最终在“任意编队”上进行复杂编队以获取“断裂”多边形集(呃,我不知道所有这些东西是什么意思)并且在事件处理器内部它最终只是调用你现有的polygon_mutable_traits::set_points
. 我已经简化了你的实现:
template <typename I>
static inline CPolygon& set_points(CPolygon& t, I f, I l)
{
t.assign(f, l);
return t;
}
然而,很明显,这种假设decltype(*I)
只是CPoint
一个错误。修复它:
template <> struct polygon_mutable_traits<CPolygon> {
template <typename I>
static inline CPolygon& set_points(CPolygon& p, I f, I l)
{
p.clear();
for (; f != l; ++f)
p.push_back({get(*f, HORIZONTAL), get(*f, VERTICAL)});
return p;
}
};
现在“一切正常”
完整演示
还将测试恢复为通用模板并在所有类型/特征之前排序:
#include <boost/polygon/polygon.hpp>
#include <cassert>
#include <fmt/ostream.h>
#include <fmt/ranges.h>
#include <list>
namespace gtl = boost::polygon;
using namespace gtl::operators;
template <typename Polygon> void test_polygon()
{
using Point = typename gtl::polygon_traits<Polygon>::point_type;
std::array const pts1{
gtl::construct<Point>(6, 0),
gtl::construct<Point>(0, 6),
gtl::construct<Point>(-6, 0),
gtl::construct<Point>(0, -6),
};
std::array const pts2{
gtl::construct<Point>(4, 0),
gtl::construct<Point>(4, 4),
gtl::construct<Point>(0, 4),
gtl::construct<Point>(0, 0),
};
Polygon poly1;
gtl::set_points(poly1, pts1.begin(), pts1.end());
Polygon poly2;
gtl::set_points(poly2, pts2.begin(), pts2.end());
#if 1
std::vector<Polygon> res;
gtl::assign(res, poly1 & poly2);
#else
std::vector<Polygon> res{poly1};
res &= poly2;
#endif
fmt::print("poly1 {}, area {}\n", poly1, gtl::area(poly1));
fmt::print("poly2 {}, area {}\n", poly2, gtl::area(poly2));
fmt::print("res {}, area {}\n", res, gtl::area(res));
}
struct CPoint {
int x;
int y;
friend std::ostream& operator<<(std::ostream& os, CPoint const& cp) {
return os << "(" << cp.x << ", " << cp.y << ")";
}
};
namespace boost::polygon {
template <> struct geometry_concept<CPoint> {
using type = point_concept;
};
template <> struct point_traits<CPoint> {
using coordinate_type = int;
static inline coordinate_type get(const CPoint& point,
orientation_2d orient)
{
return orient == HORIZONTAL ? point.x : point.y;
}
};
template <> struct point_mutable_traits<CPoint> {
using coordinate_type = int;
static inline void set(CPoint& point, orientation_2d orient, int value)
{
if (orient == HORIZONTAL) {
point.x = value;
} else {
point.y = value;
}
}
static inline CPoint construct(int x_value, int y_value)
{
return CPoint{x_value, y_value};
}
};
} // namespace boost::polygon
using CPolygon = std::list<CPoint>;
namespace boost::polygon {
template <> struct geometry_concept<CPolygon> {
using type = polygon_concept;
};
template <> struct polygon_traits<CPolygon> {
using iterator_type = CPolygon::const_iterator;
using point_type = CPoint;
using coordinate_type = typename point_traits<point_type>::coordinate_type;
static inline iterator_type begin_points(const CPolygon& t)
{
return t.begin();
}
static inline iterator_type end_points(const CPolygon& t)
{
return t.end();
}
static inline std::size_t size(const CPolygon& t) { return t.size(); }
static inline winding_direction winding(const CPolygon& t)
{
return counterclockwise_winding;
}
};
template <> struct polygon_mutable_traits<CPolygon> {
template <typename I>
static inline CPolygon& set_points(CPolygon& p, I f, I l)
{
p.clear();
for (; f != l; ++f)
p.push_back({get(*f, HORIZONTAL), get(*f, VERTICAL)});
return p;
}
};
} // namespace boost::polygon
int main()
{
test_polygon<CPolygon>();
}
印刷
poly1 {(6, 0), (0, 6), (-6, 0), (0, -6)}, area 72
poly2 {(4, 0), (4, 4), (0, 4), (0, 0)}, area 16
res {{(4, 2), (2, 4), (0, 4), (0, 0), (4, 0), (4, 2)}}, area 14
与原理图匹配:
推荐阅读
- android - 如何制作清单
可打包? - javascript - 在弹出 javascript 警报时播放音频,并在警报关闭时停止播放
- java - 开/关原则查询
- c# - 定义使用类的winForm的名称?C#
- unity3d - 统一切换平台
- java - 主/调用线程等待其他线程完成其任务的最佳方式:Thread.sleep()、Latch、CompletableFuture 还是其他?
- python - 使用列表时,然后将其随机化,如何使用 if 语句检查列表中的某个项目是否已打印?
- python - 如何使用 lambda 函数合并列表和变量
- css - 如何使用 css 拆分/切片标题或 div 并使其轻松响应?
- node.js - 如果我不舒服,我真的需要使用 React.js 吗?