如何检查boost r-tree中的任何2d点是否在给定的三角形中

r -树只作用于矩形。

如果你在r-tree中存储非矩形数据，它只会为你提供候选，你然后需要更详细地检查("验证"，"验证"，"精炼")。

整个R-tree的思想是用边界框的更简单(有效地存储和管理)几何来近似对象，毕竟。

也许boost库为此提供了一些辅助功能，但可能在rtree包之外，在几何包本身中。

如果我理解正确的话，您希望有一个包含点的r树，并检查它们中是否至少有一个与矩形或三角形相交。下面的代码展示了如何使用空间查询(rtree::query())和查询迭代器(rtree::qbegin()和rtree::qend())来实现这一点。

参见文档(http://www.boost.org/libs/geometry)，章节空间索引。

我不知道你用的是哪个编译器，所以下面的代码没有使用任何c++ 11特性。例如，在c++ 11中，您可以使用std::find_if()等带有lambda表达式的算法来代替原始循环。

#include <iostream> // to print the results
#include <vector>   // to store the points and results
// only for convenience
#include <boost/foreach.hpp>
// Boost.Geometry headers
#include <boost/geometry.hpp>
#include <boost/geometry/geometries/geometries.hpp>
#include <boost/geometry/index/rtree.hpp>
// convenient namespaces
namespace bg = boost::geometry;
namespace bgi = boost::geometry::index;
// convenient typedefs
typedef bg::model::point<double, 2, bg::cs::cartesian> point_t;
typedef bg::model::box<point_t> box_t;
typedef bg::model::ring<point_t, true, false> ring_t; // clockwise, open
typedef bgi::rtree<point_t, bgi::linear<16> > rtree_t;
int main()
{
    // prepare a container of points
    std::vector<point_t> points;
    points.push_back(point_t(0, 0));
    points.push_back(point_t(1, 1));
    points.push_back(point_t(2, 2));
    points.push_back(point_t(3, 3));
    // build a rtree
    rtree_t rtree(points.begin(), points.end());
    // prepare a triangle (not intersecting any point)
    ring_t triangle;
    bg::append(triangle, point_t(0.5, 0.6));
    bg::append(triangle, point_t(0.5, 1.5));
    bg::append(triangle, point_t(1.4, 1.5));
    // create axis-aligned bounding box of a triangle
    box_t box = bg::return_envelope<box_t>(triangle);
    // using rtree::query()
    // rather naiive approach since all points intersecting a geometry are returned
    {
        // check if at least 1 point intersecting a box was found
        std::vector<point_t> result;
        rtree.query(bgi::intersects(box), std::back_inserter(result));
        bool test = !result.empty();
        std::cout << test << std::endl;
    }
    {
        // check if at least 1 point intersecting a triangle was found
        std::vector<point_t> result;
        rtree.query(bgi::intersects(triangle), std::back_inserter(result));
        bool test = !result.empty();
        std::cout << test << std::endl;
    }
    {
        // check if at least 1 point intersecting a triangle was found
        // similar to the above but should be faster since during a spatial query
        // a box is checked and triangle only if needed
        std::vector<point_t> result;
        rtree.query(bgi::intersects(box), std::back_inserter(result));
        bool test = false;
        BOOST_FOREACH(point_t const& pt, result)
        {
            if ( bg::intersects(pt, triangle) )
            {
                test = true;
                break;
            }
        }
        std::cout << test << std::endl;
    }
    // using iterative queries - rtree::qbegin() and rtree::qend()
    // the query is stopped when the first point is found
    {
        // check if at least 1 point intersecting a box was found
        bool test = rtree.qbegin(bgi::intersects(box)) != rtree.qend();
        std::cout << test << std::endl;
    }
    {
        // check if at least 1 point intersecting a triangle was found
        bool test = rtree.qbegin(bgi::intersects(triangle)) != rtree.qend();
        std::cout << test << std::endl;
    }
    {
        // check if at least 1 point intersecting a triangle was found
        // this version should be faster than the above because a box is checked
        // during the spatial query and triangle only if needed
        bool test = false;
        // for each Point intersecting a box
        for ( rtree_t::const_query_iterator it = rtree.qbegin(bgi::intersects(box)) ;
              it != rtree.qend() ;
              ++it )
        {
            // check if this Point also intersects a triangle
            if ( bg::intersects(triangle, *it) )
            {
                test = true;
                break;
            }
        }
        std::cout << test << std::endl;
    }
}