c++ - 在四叉树中存储十亿个多边形

我需要在四叉树中存储10亿个空间多边形(使用其最小边界矩形指定)。为此，我编写了以下代码。但是，事实证明，对于10亿个点，代码运行非常缓慢。有什么方法可以改善代码，使其运行速度更快。如果是，那么有人可以帮忙吗

//---------------------------------------------------------------------------
struct MBR
{
    double xRight, xLeft, yBottom, yTop;
    MBR *zero,*first,*second,*third;
    unsigned level=0;
    vector<unsigned> result; //contains the resulting intersecting spatial ids
};
bool intersects(MBR& spatialId,MBR& mbr)
{
    if (mbr.yBottom > spatialId.yTop || mbr.yTop < spatialId.yBottom) return false;
    if (mbr.xLeft > spatialId.xRight || mbr.xRight < spatialId.xLeft) return false;
    return true;
}
//---------------------------------------------------------------------------
bool contains(MBR& spatialId,MBR& mbr)
{
    if (mbr.yBottom > spatialId.yBottom || mbr.yTop < spatialId.yTop) return false;
    if (mbr.xLeft > spatialId.xLeft || mbr.xRight < spatialId.xRight) return false;
    return true;
}
//---------------------------------------------------------------------------
bool touches(MBR& spatialId,MBR& mbr)
{
    if (    (mbr.yBottom >= spatialId.yBottom + std::numeric_limits<double>::epsilon() &&
            mbr.yBottom <= spatialId.yBottom - std::numeric_limits<double>::epsilon()) ||
            (mbr.yTop >= spatialId.yTop + std::numeric_limits<double>::epsilon() &&
            mbr.yTop <= spatialId.yTop - std::numeric_limits<double>::epsilon()))
            return true;
    if (    (mbr.xLeft >= spatialId.xLeft + std::numeric_limits<double>::epsilon() &&
            mbr.xLeft <= spatialId.xLeft - std::numeric_limits<double>::epsilon()) ||
            (mbr.xRight >= spatialId.xRight + std::numeric_limits<double>::epsilon() &&
            mbr.xRight <= spatialId.xRight - std::numeric_limits<double>::epsilon()))
            return true;
    return false;
}
//---------------------------------------------------------------------------
MBR MBR1,MBR2,MBR3,MBR4;
vector<unsigned> spatialIds; //contain 1 billion spatial identifiers which are intersected with MBR1, MBR2, MBR3, MBR4
//MBR1, MBR2, MBR3, MBR4 are again specified using their Minimum Bounding Rectangles
stack<MBR**> stackQuadTree;
MBR *root=new MBR();
(*root).yBottom=-90; (*root).yTop=90;
(*root).xLeft=-180; (*root).xRight=180;
(*root).level=0;
stackQuadTree.push(&root);

while(!stackQuadTree.empty())
{
    MBR** node=&(*stackQuadTree.front());
    if((*node)->level==50)
        break;

    (*node)->zero=new MBR(); (*node)->first=new MBR(); (*node)->second=new MBR(); (*node)->third=new MBR();
    (*node)->zero->yBottom=(*node)->yBottom; (*node)->zero->yTop=((*node)->yBottom+(*node)->yTop)/2;
    (*node)->zero->xLeft=(*node)->xLeft; (*node)->zero->xRight=((*node)->xLeft+(*node)->xRight)/2;
    (*node)->zero->level=(*node)->level+1;

    (*node)->first->yBottom=((*node)->yBottom+(*node)->yTop)/2; (*node)->first->yTop=(*node)->yTop;
    (*node)->first->xLeft=(*node)->xLeft; (*node)->first->xRight=((*node)->xLeft+(*node)->xRight)/2;
    (*node)->first->level=(*node)->level+1;

    (*node)->second->yBottom=(*node)->yBottom; (*node)->second->yTop=((*node)->yBottom+(*node)->yTop)/2;
    (*node)->second->xLeft=((*node)->xLeft+(*node)->xRight)/2; (*node)->second->xRight=(*node)->xRight;
    (*node)->second->level=(*node)->level+1;

    (*node)->third->yBottom=((*node)->yBottom+(*node)->yTop)/2; (*node)->third->yTop=(*node)->yTop;
    (*node)->third->xLeft=((*node)->xLeft+(*node)->xRight)/2; (*node)->third->xRight=(*node)->xRight;
    (*node)->third->level=(*node)->level+1;

    MBR* node=*stackQuadTree.top();
    stackQuadTree.pop();
    for(vector<MBR>::iterator itSpatialId=spatialIds.begin(),lSpatialId=spatialIds.end();itSpatialId!=lSpatialId;++itSpatialId)
    {
        if(intersects((*itSpatialId),&(*node)->zero)||contains((*itSpatialId),&(*node)->zero)||touches((*itSpatialId),&(*node)->zero))
        {
            (*node)->zero->result.push_back((*itSpatialId));
            stackQuadTree.push(*(*node)->zero);
        }

        if(intersects((*itSpatialId),&(*node)->first)||contains((*itSpatialId),&(*node)->first)||touches((*itSpatialId),&(*node)->first))
        {
            (*node)->first->result.push_back((*itSpatialId));
            stackQuadTree.push(*(*node)->first);
        }

        if(intersects((*itSpatialId),&(*node)->second)||contains((*itSpatialId),&(*node)->second)||touches((*itSpatialId),&(*node)->second))
        {
            (*node)->second->result.push_back((*itSpatialId));
            stackQuadTree.push(*(*node)->second);
        }

        if(intersects((*itSpatialId),&(*node)->third)||contains((*itSpatialId),&(*node)->third)||touches((*itSpatialId),&(*node)->third))
        {
            (*node)->third->result.push_back((*itSpatialId));
            stackQuadTree.push(*(*node)->third);
        }
    }
}

最佳答案

由于要求处理109条记录，即使使用最紧凑的表示形式，每条记录也是16字节(4 * sizeof(float))，所以我们只需要16 GiB即可将它们保存在内存中。由于我们要使用多边形的边界框进行索引，因此每个边界框可能属于多个象限。如果有很多相对较大的多边形或很多重叠，则整个四叉树的内存需求可能会大大增加。这暗示了一种迭代的方法来构建四叉树，在四叉树中，我们使用文件来存储与每个节点匹配的边界框列表以及整个树的节点信息。

四叉树图书馆

矩形表示

我们使用矩形表示两件事:

多边形的边界框(即我们的输入数据)。

象限(四叉树中的一个节点)的边界框。

我们还需要对矩形执行一些操作，例如将它们细分为多个象限，或者查找一对矩形是否相交。让我们写一个简单的类来封装它。

矩形.hpp

#pragma once

#include <cstdint>

template<typename T = float>
struct rectangle_t
{
    typedef typename T value_type;

    rectangle_t() : left(0.0f), top(0.0f), right(0.0f), bottom(0.0f) {}
    rectangle_t(value_type x1, value_type y1, value_type x2, value_type y2)
        : left(x1), top(y1), right(x2), bottom(y2)
    {
    }

    bool intersects(rectangle_t<T> const& other) const;
    bool contains(rectangle_t<T> const& other) const;
    bool touches(rectangle_t<T> const& other) const;
    rectangle_t<T> quadrant(uint32_t n) const;

    value_type left, top, right, bottom;
};

template<typename T>
inline bool rectangle_t<T>::intersects(rectangle_t<T> const& other) const
{
    return !((left > other.right)
        || (right < other.left)
        || (top > other.bottom)
        || (bottom < other.top));
}

template<typename T>
inline bool rectangle_t<T>::contains(rectangle_t<T> const& other) const
{
    return !((left >= other.left)
        || (right <= other.right)
        || (top >= other.top)
        || (bottom <= other.bottom));
}

template<typename T>
inline bool rectangle_t<T>::touches(rectangle_t<T> const& other) const
{
    return ((left == other.right)
        || (right == other.left)
        || (top == other.bottom)
        || (bottom == other.top));
}

template<typename T>
inline rectangle_t<T> rectangle_t<T>::quadrant(uint32_t n) const
{
    value_type const center_x((left + right) / 2);
    value_type const center_y((top + bottom) / 2);
    switch (n & 0x03) {
    case 0: return rectangle_t<T>(left, top, center_x, center_y);
    case 1: return rectangle_t<T>(center_x, top, right, center_y);
    case 2: return rectangle_t<T>(left, center_y, center_x, bottom);
    case 3: return rectangle_t<T>(center_x, center_y, right, bottom);
    }
    return *this; // Can't happen since we mask n
}

typedef rectangle_t<float> rectangle;

注意:虽然我们还提供了contains()和touches()方法，但我们实际上并不需要它们-如果矩形A包含矩形B，则两者也相交。同样，如果两个矩形的一对边重叠，则它们也被视为相交。

矩形序列化

下一步是开发一种将矩形存储在文件中的简单方法。我们可以使用简单的二进制文件，其布局与对象在内存中的表示方式相对应。



该文件包含单个对象序列，没有标题或其他任何内容。



由于我们希望按块处理数据，因此我们实现了提供此功能的简单读取器和写入器类。

矩形_file.hpp

#pragma once

#include "rectangle.hpp"
#include "config.hpp"

#include <fstream>
#include <string>
#include <vector>

typedef std::vector<rectangle> rectangle_vec;

class rectangle_writer
{
public:
    rectangle_writer(std::string const& file_name);
    rectangle_writer(rectangle_writer const&) = delete;

    void write_block(rectangle_vec const& block);

    uint64_t count() const { return count_; }

private:
    std::ofstream f_;
    uint64_t count_;
};

class rectangle_reader
{
public:
    rectangle_reader(std::string const& file_name);
    rectangle_reader(rectangle_reader const&) = delete;

    void read_block(rectangle_vec& block, uint64_t n = BLOCK_SIZE);

    uint64_t count() const { return count_; }
    uint64_t total_count() const { return total_count_; }
    uint64_t remaining_count() const { return total_count_ - count_; }

private:
    std::ifstream f_;
    uint64_t count_;
    uint64_t total_count_;
};

矩形_file.cpp

#include "rectangle_file.hpp"

#include <algorithm>

rectangle_writer::rectangle_writer(std::string const& file_name)
    : f_(file_name, std::ios_base::binary)
    , count_(0)
{
    if (!f_) {
        throw std::runtime_error("Unable to open index file.");
    }
}

void rectangle_writer::write_block(rectangle_vec const& block)
{
    f_.write(reinterpret_cast<char const*>(&block[0])
        , block.size() * sizeof(rectangle));
    count_ += static_cast<uint32_t>(block.size());
    if (!f_) {
        throw std::runtime_error("Read error.");
    }
}

rectangle_reader::rectangle_reader(std::string const& file_name)
    : f_(file_name, std::ios_base::binary)
    , count_(0)
{
    f_.seekg(0, std::ios::end);
    total_count_ = static_cast<uint32_t>(f_.tellg() / sizeof(rectangle));
    f_.seekg(0, std::ios::beg);
}

void rectangle_reader::read_block(rectangle_vec& block, uint64_t n)
{
    uint64_t to_read(std::min(remaining_count(), n));
    block.resize(to_read);
    if (to_read) {
        f_.read(reinterpret_cast<char*>(&block[0])
            , to_read * sizeof(rectangle));
        count_ += to_read;
        if (!f_) {
            throw std::runtime_error("Write error.");
        }
    }
}

四叉树表示

由于我们可能会使用大量节点，因此我们将所有节点保留在一个数组中，而不是使用节点索引来表示层次结构的指针。节点的索引由其在数组中的位置确定，因此我们无需显式存储该索引。

我们使用一个紧凑的结构(node_info)来表示相关的节点信息:

定义象限边界框的矩形。

子节点的4个索引的数组。 (我们使用32位索引，将我们限制在40亿个节点之内)

每个节点的匹配矩形列表存储在一个数据文件中，该文件根据节点索引(node_XXXXXXXX.dat)命名。

节点跳线

为了封装四叉树的表示并公开一个熟悉的接口(interface)以进行操作，我们创建了一个举重级类node，其中包含对quadtree的引用，节点的索引以及树中的当前级别(16字节)。此类提供了构建和使用四叉树所需的所有功能:

读写边界框。

读写匹配的矩形列表。

访问子节点，节点索引，当前级别，...

添加子节点

quadtree.hpp

#pragma once

#include "rectangle.hpp"
#include "rectangle_file.hpp"

#include <string>
#include <vector>

class quadtree
{
public:
    class node
    {
    public:
        rectangle& bounds();
        rectangle const& bounds() const;

        node child(uint32_t i) const;

        rectangle_reader reader() const;
        rectangle_writer writer() const;

        std::string rectangle_file_name() const;

        void add_child_nodes() const;

        bool is_leaf_node() const;

        uint32_t index() const;
        uint32_t level() const;

    private:
        friend class quadtree;
        node(quadtree& tree, uint32_t index, uint32_t level);

        quadtree& tree_;
        uint32_t index_;
        uint32_t level_;
    };

public:
    enum {
        ROOT_NODE_INDEX = 0
        , QUADRANT_COUNT = 4
    };

    quadtree(quadtree&& other);
    ~quadtree();

    static quadtree create(std::string const& dir_name
        , rectangle const& bounds);
    static quadtree load(std::string const& dir_name);

    void save();

    node get_node(uint32_t i = ROOT_NODE_INDEX);
    uint32_t node_count() const;

private:
    quadtree(std::string const& dir_name);
    std::string index_file_name() const;

    uint32_t add_node(rectangle const& bounds);
    void add_child_nodes(uint32_t n);

private:
    struct node_info
    {
        node_info();
        explicit node_info(rectangle const& bounds);

        rectangle bounds;
        uint32_t child[4];
    };
    typedef std::vector<node_info> node_list;

    std::string dir_name_;
    node_list nodes_;
};

inline quadtree::node::node(quadtree& tree, uint32_t index, uint32_t level)
    : tree_(tree)
    , index_(index)
    , level_(level)
{
}

inline rectangle& quadtree::node::bounds()
{
    return tree_.nodes_[index_].bounds;
}

inline rectangle const& quadtree::node::bounds() const
{
    return tree_.nodes_[index_].bounds;
}

inline quadtree::node quadtree::node::child(uint32_t i) const
{
    return node(tree_, tree_.nodes_[index_].child[i], level_ + 1);
}

inline bool quadtree::node::is_leaf_node() const
{
    return !((tree_.nodes_[index_].child[0])
        || (tree_.nodes_[index_].child[1])
        || (tree_.nodes_[index_].child[2])
        || (tree_.nodes_[index_].child[3]));
}

inline uint32_t quadtree::node::index() const
{
    return index_;
}

inline uint32_t quadtree::node::level() const
{
    return level_;
}

四叉树序列化

磁盘上的布局再次对应于对象在内存中的表示方式。



节点文件是一个简单的二进制文件，包含单个对象序列:



为了说明起见，这是在磁盘上显示具有1000个矩形的3级四叉树(根+ 2级子级)的方式:



组态

config.hpp

#pragma once

#define BLOCK_SIZE (1024 * 8)

#define QUADTREE_LOGGING 0
#define WORKQUEUE_LOGGING 0
#define PROGRESS_LOGGING 1

处理中

我们的处理基于以下观察:

仅具有根节点的树仍然是有效树。

我们可以细分任何有效的树以获得更详细的分类更深的树。

四叉树初始化

第一步是创建最简单的有效四叉树:一个只有根节点的四叉树，其中包含所有边界框。

由于我没有可以使用的任何数据，因此编写了一个简单的生成器，该生成器创建了可配置数量的随机矩形，并使用它们来初始化树。

gen_rnd_tree.cpp

#include "rectangle.hpp"
#include "rectangle_file.hpp"
#include "quadtree.hpp"
#include "config.hpp"

#include <algorithm>
#include <cstdint>
#include <chrono>
#include <iostream>
#include <random>
#include <vector>

#if defined(_WIN32)
#include <direct.h>
#endif

void generate_random(uint32_t n, quadtree& qt)
{
    rectangle_writer writer(qt.get_node().writer());

    std::vector<rectangle> buffer;
    buffer.reserve(BLOCK_SIZE);

    std::random_device rd;
    std::mt19937 gen(rd());
    std::uniform_real_distribution<rectangle::value_type> dis_x(-179.5f, 179.5f);
    std::uniform_real_distribution<rectangle::value_type> dis_y(-89.5f, 89.5f);
    std::exponential_distribution<rectangle::value_type> dis_wh(1.0f);

    for (uint32_t i(0); i < n; ++i) {
        rectangle::value_type x(dis_x(gen));
        rectangle::value_type y(dis_y(gen));
        rectangle::value_type half_w(std::min(dis_wh(gen), 10.0f) * 0.05f);
        rectangle::value_type half_h(std::min(dis_wh(gen), 10.0f) * 0.05f);
        buffer.emplace_back(x - half_w, y - half_h, x + half_w, y + half_h);

        if (buffer.size() >= BLOCK_SIZE) {
            writer.write_block(buffer);
            buffer.clear();
#if(PROGRESS_LOGGING)
            std::cout << ".";
#endif
        }
    }

    if (!buffer.empty()) {
        writer.write_block(buffer);
    }
#if(PROGRESS_LOGGING)
    std::cout << ".\n";
#endif
}

int main(int argc, char* argv[])
{
    using std::chrono::high_resolution_clock;
    using std::chrono::duration_cast;
    using std::chrono::microseconds;

    if (argc != 2) return -1;

    std::string const TREE_DIR("tree");

#if defined(_WIN32)
    _mkdir(TREE_DIR.c_str());
#else
    mkdir(TREE_DIR.c_str(), 0777);
#endif

    uint32_t const N_POLYGONS(atoi(argv[1]));

    std::cout << "Polygon count = " << N_POLYGONS << "\n";
    std::cout << "Polygon size = " << sizeof(rectangle) << " B\n";
    std::cout << "Total size = " << N_POLYGONS * sizeof(rectangle) << " B\n";

    std::cout << "\nGenerating...\n";

    high_resolution_clock::time_point t1 = high_resolution_clock::now();

    quadtree qt(quadtree::create("tree", rectangle(-180,-90,180,90)));

    generate_random(N_POLYGONS, qt);

    high_resolution_clock::time_point t2 = high_resolution_clock::now();

    std::cout << "\n";

    double dt1_us(static_cast<double>(duration_cast<microseconds>(t2 - t1).count()));
    std::cout << "Generate: " << (dt1_us / 1000.0) << " ms\n";
    std::cout << "\nDone.\n\n";
    return 0;
}

四叉树细分

函数subdivide(quadtree& qt,...)使用堆栈执行树的深度优先遍历。我们可以使用队列进行广度优先遍历，但是一旦我们深入到树中，队列就会变得非常大。

我们反复从堆栈顶部消耗节点，尝试对其进行细分，然后将其子节点推入堆栈。

函数subdivide(quadtree::node const& qtn, ...)决定是否细分给定节点，并为copy_elements(...)准备参数。一旦节点被细分，并且所有匹配的矩形都已移动到子节点，我们将清除给定节点的矩形文件。

函数copy_elements(...)从父节点读取匹配矩形的列表，并将其复制到边界相交的所有子节点。

#include "rectangle.hpp"
#include "rectangle_file.hpp"
#include "quadtree.hpp"

#include <algorithm>
#include <cstdint>
#include <chrono>
#include <iostream>
#include <stack>

using std::chrono::high_resolution_clock;
using std::chrono::duration_cast;
using std::chrono::microseconds;

void copy_elements(rectangle_reader& reader, quadtree::node child_node[4])
{
    rectangle target[quadtree::QUADRANT_COUNT] {
         child_node[0].bounds()
        , child_node[1].bounds()
        , child_node[2].bounds()
        , child_node[3].bounds()
    };

    rectangle_writer writer[quadtree::QUADRANT_COUNT] {
        child_node[0].writer()
        , child_node[1].writer()
        , child_node[2].writer()
        , child_node[3].writer()
    };

    rectangle_vec buffer_in;
    buffer_in.reserve(BLOCK_SIZE);
    rectangle_vec buffer_out[quadtree::QUADRANT_COUNT];
    for (auto& buffer : buffer_out) {
        buffer.reserve(BLOCK_SIZE);
    }

#if(PROGRESS_LOGGING)
    for (; reader.remaining_count(); std::cout << ".") {
#else
    for (; reader.remaining_count();) {
#endif
        reader.read_block(buffer_in, BLOCK_SIZE);

        for (auto const& rect : buffer_in) {
            for (uint32_t i(0); i < quadtree::QUADRANT_COUNT; ++i) {
                if (target[i].intersects(rect)) {
                    buffer_out[i].push_back(rect);
                }
            }
        }

        for (uint32_t i(0); i < quadtree::QUADRANT_COUNT; ++i) {
            if (!buffer_out[i].empty()) {
                writer[i].write_block(buffer_out[i]);
                buffer_out[i].clear();
            }
        }
    }
}

// Return true if we should subdivide further, false otherwise
bool subdivide(quadtree::node const& qtn, uint32_t max_depth, uint32_t max_leaf_size)
{
    bool at_max_depth(qtn.level() >= max_depth);
    std::cout << (at_max_depth ? "Skipping" : "Subdividing")
        << " node #" << qtn.index() << " @ level " << qtn.level() << ".\n";
    if (at_max_depth) {
        return false;
    } else {
        // Make sure reader goes out of scope before writer is created.
        rectangle_reader reader(qtn.reader());
        if (qtn.is_leaf_node()) {
            if (reader.total_count() <= max_leaf_size) {
                std::cout << "Leaf node (" << reader.total_count() << ").\n";
                return false;
            }
            qtn.add_child_nodes();
        } else if (!reader.total_count()) {
            std::cout << "Branch node.\n";
            return (qtn.level() < max_depth);
        }

        quadtree::node child_nodes[4] {
            qtn.child(0)
            , qtn.child(1)
            , qtn.child(2)
            , qtn.child(3)
        };

        high_resolution_clock::time_point t1(high_resolution_clock::now());

        copy_elements(reader, child_nodes);

        high_resolution_clock::time_point t2(high_resolution_clock::now());
        double dt_us(static_cast<double>(duration_cast<microseconds>(t2 - t1).count()));

        std::cout << ". (" << reader.count() << ") " << (dt_us / 1000.0) << " ms\n";
    }

    rectangle_writer writer(qtn.writer()); // Clean the file

    std::cout << "Branch node.\n";
    return (qtn.level() < max_depth);
}

void subdivide(quadtree& qt, uint32_t max_depth, uint32_t max_leaf_size)
{
    std::stack<quadtree::node> work;
    work.emplace(qt.get_node());
    std::size_t max_queue_size(work.size());

    for (; !work.empty();) {
        quadtree::node node(work.top());
        work.pop();

        if (subdivide(node, max_depth, max_leaf_size)) {
            for (uint32_t i(0); i < quadtree::QUADRANT_COUNT; ++i) {
                work.emplace(node.child(i));
            }
        }
        max_queue_size = std::max(max_queue_size, work.size());
#if(WORKQUEUE_LOGGING)
        std::cout << "* Work queue " << work.size() << " (" << max_queue_size << ").\n";
#endif
    }
}

int main(int argc, char* argv[])
{
    uint32_t const MAX_DEPTH(atoi((argc >= 2) ? argv[1] : "2"));
    uint32_t const MAX_LEAF_SIZE(atoi((argc >= 3) ? argv[2] : "100"));

    std::cout << "Loading...\n";

    high_resolution_clock::time_point t1(high_resolution_clock::now());

    quadtree qt(quadtree::load("tree"));

    if (qt.node_count() < 1) {
        std::cerr << "Invalid tree.\n";
        return -1;
    }

    std::cout << "Building tree (max_depth=" << MAX_DEPTH
        << ", max_leaf_size=" << MAX_LEAF_SIZE << ")...\n";

    high_resolution_clock::time_point t2(high_resolution_clock::now());

    subdivide(qt, MAX_DEPTH, MAX_LEAF_SIZE);

    high_resolution_clock::time_point t3(high_resolution_clock::now());

    std::cout << "\n";

    std::cout << "Tree completed (" << qt.node_count() << " nodes).\n\n";

    double dt1_us(static_cast<double>(duration_cast<microseconds>(t2 - t1).count()));
    double dt2_us(static_cast<double>(duration_cast<microseconds>(t3 - t2).count()));
    std::cout << "Load: " << (dt1_us / 1000.0) << " ms\n";
    std::cout << "Process: " << (dt2_us / 1000.0) << " ms\n";
    std::cout << "\nDone.\n\n";
    return 0;
}

用法

gen_rnd_tree <count>-在根节点中生成带有计数矩形的树。

build_tree <max_depth> <max_node_size>-细分具有多个max_node_size个矩形的节点，最大深度为max_depth(根在深度0)。

样本输出

Polygon count = 1000
Polygon size = 16 B
Total size = 16000 B

Generating...
.

Generate: 0 ms

Done.

Loading...
Building tree (max_depth=5, max_leaf_size=100)...
Subdividing node #0 @ level 0.
.. (1000) 15.624 ms
Branch node.
Subdividing node #4 @ level 1.
.. (288) 15.625 ms
Branch node.
Subdividing node #8 @ level 2.
Leaf node (71).
Subdividing node #7 @ level 2.
Leaf node (59).
Subdividing node #6 @ level 2.
Leaf node (81).
Subdividing node #5 @ level 2.
Leaf node (77).
Subdividing node #3 @ level 1.
.. (212) 0 ms
Branch node.
Subdividing node #12 @ level 2.
Leaf node (46).
Subdividing node #11 @ level 2.
Leaf node (55).
Subdividing node #10 @ level 2.
Leaf node (55).
Subdividing node #9 @ level 2.
Leaf node (58).
Subdividing node #2 @ level 1.
.. (260) 15.626 ms
Branch node.
Subdividing node #16 @ level 2.
Leaf node (68).
Subdividing node #15 @ level 2.
Leaf node (69).
Subdividing node #14 @ level 2.
Leaf node (71).
Subdividing node #13 @ level 2.
Leaf node (53).
Subdividing node #1 @ level 1.
.. (240) 0 ms
Branch node.
Subdividing node #20 @ level 2.
Leaf node (68).
Subdividing node #19 @ level 2.
Leaf node (60).
Subdividing node #18 @ level 2.
Leaf node (62).
Subdividing node #17 @ level 2.
Leaf node (50).

Tree completed (21 nodes).

Load: 0 ms
Process: 109.377 ms

Done.

当尝试对108k个矩形(64k记录的BLOCK_SIZE)进行尝试时，用max_depth为6的树构建大约花了45秒钟。使用内存算法可以轻松处理105个项目。

该程序在整个过程中仅使用了很少的RAM MiB。

进一步改进

错误检查，单元测试(我在这里无法放入代码)。

将元数据添加到矩形文件，以便您可以找到它们匹配的多边形。我将记录的大小保持16字节的倍数(对齐)。

并行化

Mutex同步quadtree，互斥锁同步stack，线程池运行for (; !work.empty();) { ...循环。

由于我们仅在子节点细分后才推送子节点，因此每个线程将始终在树的隔离部分上工作。

混合方法

磁盘I / O成为此类处理的瓶颈。一旦到达足够小的节点(例如105-106个条目)，我们就可以在内存中方便地处理数据。

使用此算法中的节点列表作为查找以加载“小型”内存四叉树。实现缓存机制，以保持最近查找的“小”四叉树可用。

向量化(也许可以加快匹配大量矩形的速度)。

关于c++ - 在四叉树中存储十亿个多边形，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/36454103/