数据结构

        声明八叉树中简单点OctreePoint的数据类型、初始化方式、以及OctreePoint位置position的两个 *** 作函数。

#ifndef OctreePoint_H
#define OctreePoint_H

#include "Vec3.h"

// Simple point data type to insert into the tree.
// Have something with more interesting behavior inherit  from this in order to store other attributes in the tree.
// 要插入八叉树中简单点的数据类型。

// 继承一些具有更有趣行为的东西，以便在树中存储其他属性。

class OctreePoint {
	Vec3 position; 
public:
	OctreePoint() { }//初始化OctreePoint
	OctreePoint(const Vec3& position) : position(position) { } //初始化OctreePoint，并给OctreePoint的position赋值
	inline const Vec3& getPosition() const { return position; }  //返回/获取位置数据
	inline void setPosition(const Vec3& p) { position = p; }        //设置位置数据
};

#endif

        定义一个用于计时的函数stopwatch()，时间单位为秒

#ifndef _WIN32
#include 

double stopwatch()
{
	struct timeval time;
	gettimeofday(&time, 0 );
	return 1.0 * time.tv_sec + time.tv_usec / (double)1e6; //单位为秒
}

#else

#include 
double stopwatch() 
{
	unsigned long long ticks;
	unsigned long long ticks_per_sec;
	QueryPerformanceFrequency( (LARGE_INTEGER *)&ticks_per_sec);
	QueryPerformanceCounter((LARGE_INTEGER *)&ticks);
	return ((float)ticks) / (float)ticks_per_sec;
}

#endif

        声明一个八叉树的类，基本组成为:当前节点的位置origin、当前节点对应空间范围halfDimension、当前节点的子节点children、当前节点的数据data。

        然后递归定义了八叉树当前节点的八个子节点的八叉树；以及子节点序号查询函数和某一空间范围框内八叉树节点(当前节点、子节点、子子节点。

。

。

。

)的查询方法

#ifndef Octree_H
#define Octree_H

#include 
#include 
#include "OctreePoint.h"

namespace brandonpelfrey {

	/**!
	 *定义一个八叉树的类，其组成包括:
	 	当前节点中心位置origin、当前节点对应空间尺寸halfDimension、
	 	当前节点的子节点children、当前节点的数据data
	 */
	class Octree {
		// Physical position/size. This implicitly defines the bounding box of this node   
		// 物体位置和大小，间接定义了当前节点框边界
  		Vec3 origin;         //! The physical center of this node  该节点的物理中心
		Vec3 halfDimension;  //! Half the width/height/depth of this node 该节点的半宽/高/深

		//这棵树最多有八个子节点，分别可以额外存储一个点，不过在许多应用程序中，叶子将存储数据。

		// The tree has up to eight children and can additionally store 
		// a point, though in many applications only, the leaves will store data.
		Octree *children[8]; //! Pointers to child octants   指向八叉树子节点的指针
		OctreePoint *data;   //! Data point to be stored at a node  一个指针--指向存储在一个八叉树节点中的数据

		/*
				Children follow a predictable pattern to make accesses simple.
				八叉树子节点遵循可预测的模式以使访问变得简单。

				Here, - means less than 'origin' in that dimension, + means greater than.
				这里，- 表示小于该维度(坐标)的“原点”，+ 表示大于。

				child:	0 1 2 3 4 5 6 7  //从下面xyz的+-表示中可以看出这个八个子节点的空间分布
				x:      - - - - + + + +
				y:      - - + + - - + +
				z:      - + - + - + - +
		 */

		public:
		Octree(const Vec3& origin, const Vec3& halfDimension) //初始化一个八叉树
			: origin(origin), halfDimension(halfDimension), data(NULL) {
				// Initially, there are no children
				for(int i=0; i<8; ++i) 
					children[i] = NULL;
			}

		Octree(const Octree& copy) //从另一个八叉数复制得到一个八叉树
			: origin(copy.origin), halfDimension(copy.halfDimension), data(copy.data) {

			}

		~Octree() {
			// Recursively destroy octants //递归删除子节点
			for(int i=0; i<8; ++i) 
				delete children[i];
		}

		// 确定point在该八叉树的哪一个子节点空间内
		// Determine which octant of the tree would contain 'point'
		int getOctantContainingPoint(const Vec3& point) const {
			int oct = 0;
			if(point.x >= origin.x) oct |= 4; //"|="位或 *** 作
			if(point.y >= origin.y) oct |= 2;
			if(point.z >= origin.z) oct |= 1;
			return oct;//这个值应该是0、1、2、3、4、5、6、7之中的一个
		}

		//检查是否是叶子节点(叶子节点--没有子节点的八叉树节点)
		bool isLeafNode() const {
			// This is correct, but overkill--这是正确的，但矫枉过正. See below.
			/*
				 for(int i=0; i<8; ++i)
				 if(children[i] != NULL) 
				 return false;
				 return true;
			 */

			// We are a leaf if we have no children. Since we either have none, or  all eight, it is sufficient to just check the first.
			// 如果没有子节点，就是一片叶子节点。
 由于要么没有，要么全部都没有，所以只检查第一个就足够了。

			return children[0] == NULL;
		}

		//插入一个简单数据点，即将简单数据点插入到八叉树中
		void insert(OctreePoint* point) {
			// If this node doesn't have a data point yet assigned  and it is a leaf, then we're done!
			// 如果八叉树当前节点还没有分配一个数据点并且它是一个叶子节点，那么我们就完成了----把该简单数据点赋值到当前节点即可
			if(isLeafNode()) { //是否是叶子节点----没有子节点
				if(data==NULL) { //当是叶子节点并且没有被赋值时，将简单数据点赋值给它
					data = point;
					return;
				} else { 
					// We're at a leaf, but there's already something here
					// We will split this node so that it has 8 child octants and then insert the old data that was here, along with this new data point 
					// Save this data point that was here for a later re-insert
					// 是叶子节点并且已经被赋值时
					// 将分割这个叶子节点，使其有 8 个子节点，然后插入这里的旧数据以及这个新数据点
					// 保存此数据点，以便稍后重新插入
					OctreePoint *oldPoint = data; //保存叶子节点的数据
					data = NULL;

					// Split the current node and create new empty trees for each  child octant.
					// 拆分当前叶子节点，并为每个子节点创建新的空八叉树。

					for(int i=0; i<8; ++i) {
						// Compute new bounding box for this child 为子节点创建新的边界框
						Vec3 newOrigin = origin;
						//&运算符、三目运算符
						newOrigin.x += halfDimension.x * (i&4 ? .5f : -.5f);//i为4时
						newOrigin.y += halfDimension.y * (i&2 ? .5f : -.5f);//i为2、6时
						newOrigin.z += halfDimension.z * (i&1 ? .5f : -.5f);//i为1、3、5、7时
						children[i] = new Octree(newOrigin, halfDimension*.5f);
					}

					// Re-insert the old point, and insert this new point 重新插入旧点，并插入这个新点
					// (We wouldn't need to insert from the root, because we already know it's guaranteed to be in this section of the tree)
					//（我们不需要从根插入，因为我们已经知道它保证在八叉树的某一个子节点对应的空间内）
					// 这里对子节点进行插入“简单数据点”操作，观察children[]里面得到的子节点的序号
					children[getOctantContainingPoint(oldPoint->getPosition())]->insert(oldPoint);
					children[getOctantContainingPoint(point->getPosition())]->insert(point);
				}
			} else {
				// We are at an interior node. Insert recursively into the  appropriate child octant
				// 当八叉树有子节点时，根据需要插入的简单数据点得到子节点的序号，并将该简单数据点插入其中
				int octant = getOctantContainingPoint(point->getPosition());
				children[octant]->insert(point);
			}
		}

		// This is a really simple routine for querying the tree for points within a bounding box defined by min/max points (bmin, bmax)
		// 这是一个非常简单的例程，用于查询八叉树中（bmin，bmax）定义的边界框内的八叉树节点
		// All results are pushed into 'results'
		void getPointsInsideBox(const Vec3& bmin, const Vec3& bmax, std::vector& results) {
			// If we're at a leaf node, just see if the current data point is inside the query bounding box
			// 如果我们在叶子节点，只需查看当前叶子节点是否在查询边界框内
			if(isLeafNode()) {
				if(data!=NULL) {
					const Vec3& p = data->getPosition();//当前叶子节点的数据
					if(p.x>bmax.x || p.y>bmax.y || p.z>bmax.z) return;
					if(p.x
 我们将检查这八个子节点是否位于查询边界框内
				for(int i=0; i<8; ++i) {
					// Compute the min/max corners of this child octant 计算这个某个子节点的最小/最大范围
					Vec3 cmax = children[i]->origin + children[i]->halfDimension;
					Vec3 cmin = children[i]->origin - children[i]->halfDimension;

					// If the query rectangle is outside the child's bounding box,  then continue
					// 如果查询矩形在子节点的边界框之外，则继续（其他子节点或向上一级节点搜索）
					if(cmax.xbmax.x || cmin.y>bmax.y || cmin.z>bmax.z) continue;

					// At this point, we've determined that this child is intersecting  the query bounding box
					// 至此，我们已经确定这个子节点在查询边界框内，进一步进行搜索(这个子节点是否有子子节点。
。
。
。
)
					children[i]->getPointsInsideBox(bmin,bmax,results);
				} 
			}
		}

	};

}
#endif

        创建了一组随机点，比较使用八叉树查询和使用强制/蛮力查询所需要的时间；并指出，当在查询相对接近整个八叉树空间大小的情况下，八叉树实际上会比蛮力/强制查询个点慢一点！

#include 
#include 
#include 
#include 

#include "Octree.h"
#include "Stopwatch.h"

using namespace brandonpelfrey;

// Used for testing
std::vector points;
Octree *octree;
OctreePoint *octreePoints;
Vec3 qmin, qmax;

float rand11() // Random number between [-1,1] //返回[-1,1]之间的随机数字
{ return -1.f + (2.f*rand()) * (1.f / RAND_MAX); }

//创建随机向量，其中组成元素在[-1,1]范围内
Vec3 randVec3() // Random vector with components in the range [-1,1]
{ return Vec3(rand11(), rand11(), rand11()); }

// Determine if 'point' is within the bounding box [bmin, bmax] //确定点是否在框里面
bool naivePointInBox(const Vec3& point, const Vec3& bmin, const Vec3& bmax) {
	return 
		point.x >= bmin.x &&
		point.y >= bmin.y &&
		point.z >= bmin.z &&
		point.x <= bmax.x &&
		point.y <= bmax.y &&
		point.z <= bmax.z;
}

void init() {
	// Create a new Octree centered at the origin with physical dimension 2x2x2
	// 创建一个中心在原点的八叉树，物理尺寸为2*2*2
	octree = new Octree(Vec3(0,0,0), Vec3(1,1,1));//初始化在原点附近，为得是接下来创建随机点直观些

	// Create a bunch of random points 创建一组随机点
	const int nPoints = 1 * 1000 * 1000;
	for(int i=0; i

	因为下一个数据再上一个数据还没输出完毕，还在输出缓冲区中时，
	下一个printf就把另一个数据加入输出缓冲区，结果冲掉了原来的数据，出现输出错误。
 
	在 prinf（）；后加上fflush(stdout); 强制马上输出，避免错误。
*/
	printf("Created %ld points\n", points.size()); fflush(stdout);

	// Insert the points into the octree  在八叉树中插入点
	octreePoints = new OctreePoint[nPoints];
	for(int i=0; iinsert(octreePoints + i);//插入简单数据点
	}
	printf("Inserted points to octree\n"); fflush(stdout);

	// Create a very small query box. The smaller this box is the less work the octree will need to do. 
	// 创建一个非常小的查询框。
 这个框越小，八叉树需要做的工作就越少。

	// This may seem like it is exagerating the benefits, but often, we only need to know very nearby objects.
	// 这似乎夸大了好处，但通常，我们只需要知道非常近的物体。

	qmin = Vec3(-.05,-.05,-.05);
	qmax = Vec3(.05,.05,.05);

	// Remember: In the case where the query is relatively close to the size of the whole octree space, the octree will
	// actually be a good bit slower than brute forcing every point!
	// 请记住：在查询相对接近整个八叉树空间大小的情况下，八叉树实际上会比蛮力/强制查询个点慢一点！
}

// Query using brute-force 使用强制查询框内的点
void testNaive() {
	double start = stopwatch();//用于计时

	std::vector results;
	for(int i=0; i results;
	octree->getPointsInsideBox(qmin, qmax, results);

	double T = stopwatch() - start;
	printf("testOctree found %ld points in %.5f sec.\n", results.size(), T);
}


int main(int argc, char **argv) {
	init();
	testNaive();
	testOctree();

	return 0;
}

        根据下面对比，发现查询框较小是八叉树查询很快；当查询框接近整个空间时，八叉树查询会慢很多；而蛮力查询所需时间一直很稳定，甚至当查询空间小于一般时，会降低

（1）查询框大小为

//main.cpp的63、64行
	qmin = Vec3(-.05,-.05,-.05);
	qmax = Vec3(.05,.05,.05);

编译并运行，得到八叉树查询用时：0.00003 sec.蛮力查询用时0.00711sec.

meng@meng:~/SimpleOctree$ make
g++ main.cpp -O3 -o octree
meng@meng:~/SimpleOctree$ ./octree 
Created 1000000 points
Inserted points to octree
testNaive found 150 points in 0.00711 sec.
testOctree found 150 points in 0.00003 sec.

 （2）查询框大小为

	qmin = Vec3(-.5,-.5,-.5);
	qmax = Vec3(.5,.5,.5);

        删除可执行文件octree，重新编译

Created 1000000 points
Inserted points to octree
testNaive found 150 points in 0.00726 sec.
testOctree found 150 points in 0.00004 sec.

 （3）查询框大小为

	qmin = Vec3(-.9,-.9,-.9);
	qmax = Vec3(.9,.9,.9);

meng@meng:~/SimpleOctree$ ./octree 
Created 1000000 points
Inserted points to octree
testNaive found 729239 points in 0.00554 sec.
testOctree found 729239 points in 0.09651 sec.

        更多对比自己验证啦

代码来源：https://github.com/brandonpelfrey/SimpleOctree

@meng