游戏AI寻路——八叉树+A*寻路

利用八叉树的空中寻路

你有思考过在空中如何进行寻“路”吗?来想象一个的场景:飞机从空中基地出发,要避开许多空中建筑,最终到达目的地。这种情况下的寻路是没有路面的,寻路物体的移动方向也比较自由,这该怎么寻呢?

如果我们只是在一个平面进行寻路,我们可以直接用A*寻路,铺好一个地面网格,这样就可以在网格点上设置目标点来寻路了。假设我们要在一个 \(500\times500\) 大小的网格寻路,就算一个单位设置一个网格点,那就要 \(500 \times 500 = 25,0000\) 这么多个点,不过倒也是不能接受。

现在我们算上“领空”,就算取100得到的数值 \(500 \times 500 \times 100\) 也是挺大的……有办法减少结点保证网格连接合理吗?如果解决这两个问题,也不是不能继续使用A*寻路。

欸,这就可以通过八叉树来实现!

注意:文中代码部分有些地方会用省略号,表示「对应部分内容与之前一样,不需要修改」,是为了突出重点内容。如需要完整代码,文末会给出。

寻路中八叉树的作用

利用八叉树的寻路,并不是说要用八叉树做一个像A*那样的寻路算法,而是利用它来生成寻路区域。可以认为它是另一种寻路网格,八叉树最终生成的会比之前我们想的那种笨方法的结点更少,在八叉树生成的网格里我们依然可以使用原本的寻路算法。

PS:八叉树还有其它的正经工作,比如碰撞检测,对引擎开发感兴趣的同学也可以去了解一下。

生成寻路网格

1. 八叉树结点

现在就要看看如何用八叉树来生成寻路结点了。先说说八叉树吧,八叉树本身并不复杂,它说的是这么一个结构:

所谓“一尺之棰,日取其半,万世不竭”,难道要一直分下去吗?我们可以给它设置一个最小尺寸来限制,只有当前方块尺寸比最小尺寸大时才分裂,至此,我们可以初步构建八叉树结点:

using System.Collections;
using System.Collections.Generic;
using UnityEngine;

public class MyOctreeNode
{
    private const float MIN_CUBE_SIZE = 1f; // 最小方格尺寸
    public MyOctreeNode Parent{ get; set; } //父结点 
    public MyOctreeNode[] Children; //子结点
    public Bounds NodeCube; //用包围盒作为结点方块,方便后续检测

    public MyOctreeNode(Bounds nodeCube, MyOctreeNode parent)
    {
        Parent = parent;
        NodeCube = nodeCube;
    }
    public void Divide()
    {
        //因为是正方体,所以用一条边来判断尺寸即可
        if(NodeCube.size.x >= MIN_CUBE_SIZE) 
        {
            // 子方块的半尺寸, 用半尺寸是因为构建Bounds需要
            float childHalfSize = NodeCube.size.x / 4;
            if (Children == null)
                Children = new MyOctreeNode[8];
            Vector3 offset; //子结点偏移
            for(int i = 0; i < 8; ++i)
            {
                //待补充
                var childBounds = new Bounds();
                //

                if(Children[i] == null)
                    Children[i] = new MyOctreeNode(childBounds, this);
                Children[i].Divide(); // 每个子结点继续分裂
            }
        }
    }
}

子结点的方块该怎么布置呢?简单分析下位置关系就可以看出来:

每个子方块对于原本方块中心的各轴的偏移量都是原本边长的 \(\frac{1}{4}\),无非是 \(+\frac{1}{4}、-\frac{1}{4}\) 的差别。但好在,我们不关心子结点的顺序(也就是说在数组中这八个方块谁先谁后都无所谓),那么这8种正负号的组合方案可以通过对0~7的数取二进制的3个位来得到(下图0 ~ 7是乱序的,只是为了对照):

当然,如果你觉得不够直观,也可以用数组记录这8个情况再遍历赋值,这里就只是图省个数组而已。那就用上述方法完善一下Divide方法:

public void Divide()
{
    //因为是正方体,所以用一条边来判断尺寸即可
    if(NodeCube.size.x >= MIN_CUBE_SIZE) 
    {
        // 子方块的半尺寸, 用半尺寸是因为构建Bounds需要
        float childHalfSize = NodeCube.size.x / 4;
        if (Children == null)
            Children = new MyOctreeNode[8];
        Vector3 offset; //子节点偏移
        for(int i = 0; i < 8; ++i)
        {
            //0~7的二进制位结构恰好满足我们所需要的组合形式
            offset.x = (1 & i) != 0 ? childHalfSize : -childHalfSize; //取二进制第0位
            offset.y = (2 & i) != 0 ? childHalfSize : -childHalfSize; //取二进制第1位
            offset.z = (4 & i) != 0 ? childHalfSize : -childHalfSize; //取二进制第2位
            var childBounds = new Bounds(NodeCube.center + offset, 2 * childHalfSize * Vector3.one);
            
            if(Children[i] == null)
                Children[i] = new MyOctreeNode(childBounds, this);
            Children[i].Divide(); // 每个子节点继续分裂
        }
    }
}

为了方便观察结果,再在类中添个用于绘制方块的函数,当它在OnDrawGizmos中调用时就可以看到方块了:

//isSeeOne为true,则只查看分裂后的一个,否则查看所有分裂后的方块
public void Draw(bool isSeeOne)
{
    Gizmos.color = Color.green;
    Gizmos.DrawWireCube(NodeCube.center, NodeCube.size);
    if (Children == null)
        return;
    foreach(var c in Children)
    {
        c.Draw(isSeeOne);
        if(isSeeOne)
        {
            break;
        }
    }
}

为了方便在Unity中使用,我们创建一个继承了MonoBehaviour的类MyOctreeBuilder,并将它挂在一个边长为8的Cube上:

using System.Collections;
using System.Collections.Generic;
using UnityEngine;

public class MyOctreeBuilder : MonoBehaviour
{
    public bool isSeeOne = true; //只看分裂后的一个
    private MyOctreeNode node;
    private void Awake()
    {
        //用Cube本身的包围盒做为起始尺寸进行划分
        node = new MyOctreeNode(GetComponent<Renderer>().bounds, node);
        node.Divide();
    }

    private void OnDrawGizmos()
    {
        if (Application.isPlaying)
        {
            node.Draw(isSeeOne);
        }
    }
}

我们设置的最小尺寸为1,从8减半到1,一共要3次,划分出的方块数符合预期。

2. 根结点

那要如何设置包围盒才能让它刚好能包围我的场景呢,总不能拿Cube去自己试吧?欸,好在Unity的Bounds类有个可以帮助我们的方法:

Encapsulate方法可以让包围盒自行扩大以容纳下传进来的包围盒。所以我们让一个包围盒把场景中的所有物体都容纳进去,这样就能得到足够大的包围盒了。我们新建一个MyOctree类:

using System.Collections;
using System.Collections.Generic;
using UnityEngine;

public class MyOctree
{
    public MyOctreeNode RootNode;
    public MyOctree(GameObject[] allObjects)
    {
        var baseCube = new Bounds();
        foreach(var o in allObjects)
        {
            baseCube.Encapsulate(o.GetComponent<Collider>().bounds);
        }
        //选取最长的一条边来作为正方体的边长,并将包围盒改成正方体
        //这里为了更好设置包围盒,同样记录半尺寸
        var cubeHalfSize = 0.5f * Mathf.Max(baseCube.size.x, baseCube.size.y, baseCube.size.z) * Vector3.one;
        baseCube.SetMinMax(baseCube.center - cubeHalfSize, baseCube.center + cubeHalfSize);

        RootNode = new MyOctreeNode(baseCube, null);
        RootNode.Divide();
    }
}

顺便也改改MyOctreeBuilder脚本,让它画出八叉树,而不是单一节点:

public class MyOctreeBuilder : MonoBehaviour
{
    public GameObject[] objects; //需要包含的物体
    public bool isSeeOne = true; //只看分裂后的一个
    private MyOctree myOctree; //八叉树
    
    private void Awake()
    {
        myOctree = new MyOctree(objects);
    }

    private void OnDrawGizmos()
    {
        if (Application.isPlaying)
        {
            myOctree.RootNode.Draw(isSeeOne);
        }
    }
}

随便在场景里摆了几个立方体,最终生成的最大包围盒能将它们都裹住:

至此,准备工作完成。

3. 剔除不必要的结点(关键)

仅是不断分裂生成小方块,那最终不还是和我们开头的笨方法一样吗(会有一堆密密麻麻的点)?我们可以注意到,其实有一些方块没必要继续分裂下去。分裂行为其实是有目的的:检测出哪里有障碍物。一个大方块不断分裂变小,就是更进一步定位内部障碍物位置的过程,如果它一开始就没碰到什么障碍物,那也没必要分裂了。

我们需要对先前几个类中的内容稍加修改:

  1. MyOctreeNode类的Divide方法中分裂前要进行一些条件判断:
    public void Divide(Collider collider)
    {
        //因为是正方体,所以用一条边来判断尺寸即可
        if(NodeCube.size.x >= MIN_CUBE_SIZE) 
        {
            // 子方块的半尺寸, 用半尺寸是因为构建Bounds需要
            float childHalfSize = NodeCube.size.x / 4;
            if (Children == null)
                Children = new MyOctreeNode[8];
            Vector3 offset; //子节点偏移
            for(int i = 0; i < 8; ++i)
            {
                //0~7的二进制位结构恰好满足我们所需要的组合形式
                offset.x = (1 & i) != 0 ? childHalfSize : -childHalfSize;
                offset.y = (2 & i) != 0 ? childHalfSize : -childHalfSize;
                offset.z = (4 & i) != 0 ? childHalfSize : -childHalfSize;
                var childBounds = new Bounds(NodeCube.center + offset, 2 * childHalfSize * Vector3.one);
                if(Children[i] == null)
                    Children[i] = new MyOctreeNode(childBounds, this);
                /*
                进一步分裂前,先判断一下有没有遇到障碍物,没有就不要继续分裂了;
                也可以再附带添加些其它检测条件,比如obj.layer等
                */
                if(childBounds.Intersects(collider.bounds))
                {
                    Children[i].Divide(collider); // 每个子节点继续分裂
                }
            }
        }
    }
    
  2. MyOctree类初始化时具体分裂:
    public MyOctree(GameObject[] allObjects)
    {
        var baseCube = new Bounds();
        foreach(var o in allObjects)
        {
            baseCube.Encapsulate(o.GetComponent<Collider>().bounds);
        }
        //选取最长的一条边来作为正方体的边长,并将包围盒改成正方体
        //这里为了更好设置包围盒,同样记录半尺寸
        var cubeHalfSize = 0.5f * Mathf.Max(baseCube.size.x, baseCube.size.y, baseCube.size.z) * Vector3.one;
        baseCube.SetMinMax(baseCube.center - cubeHalfSize, baseCube.center + cubeHalfSize);
    
        RootNode = new MyOctreeNode(baseCube, null);
        foreach(var o in allObjects) //具体分裂
        {
            //有碰撞体的物体才有检测的必要
            if(o.TryGetComponent(out Collider collider))
            {
                RootNode.Divide(collider);
            }	
        }
    }
    

我们是在进一步分裂前制止分裂的,算是一种前剪枝策略;相对的,如果是在所有结点都生成好后,再删掉不必要的结点,那就是后剪枝策略。考虑到场景可能会很大,前剪枝的策略明显会更好些。

让我们看看修改后的效果:

八叉树替代网格的关键就在于此:存在障碍物的地方才会有密集的结点,空旷的地方倒没什么结点。

这其实很符合人的自然智慧:首先我们要明白结点多意味着什么,这其实意味着能更精细的寻路。在有障碍物的地方,我们就得小心避障、“步步为营”,所以需要更多结点细化落脚点;而空旷的地方就不用这样绕来绕去,直接“两点一线”就够了。

4. 连接成网

结点已经全部划分出来了,那如何连接成网格呢?我们可以自然而然想到两种做法(当然,可以有其它做法):

  1. 父子相连:每个结点都与它的父结点和子结点相连(如果有的话)
  2. 全连接:每个结点和其它结点依次相连

这两种朴素的做法其实已经反映出了连接需要考虑的问题:

首先,这两种做法都可以算是对的。它们都能保证整个网络是连通的,也就说,在这两种方法构建的网格下,我们总可以找到路径从一个结点到另外一个结点。

对于第一种做法,它连接成网所构建的边的数量明显比第二种来得少。但很显然,由于边的数量过少,实际的路径选择也会很少,即使是去不同的地方,走出的路径也是大同小异,并且还会出现绕远路的情况。

而第二种就是另一个极端了,它所构建的网格连通性极高,两点之间通常都含有着丰富的路径选择,但需要存储的边实在是太多了。

哪种方案更好?这是根据实际情况调整的。

这里我们采用一种第二种策略,但有一点要注意:我们应该把全是障碍物的结点排除掉,因为它们所在的位置已经没有行走的余地了。

现在就来实现一下:

  1. 我们准备一个枚举,来区分结点的类型(也方便后续拓展),暂时就分两类结点:通常、障碍(针对最小障碍)。并在分裂过程中判别哪些是障碍。根据我们的分裂逻辑,可以清楚地想到:只要仍需分裂的最小结点才是最小障碍:

    public enum NodeType
    {
        Normal, Obstacles,
    }
    public class MyOctreeNode
    {
        //……
        public Bounds NodeCube; //用包围盒作为结点方块,方便后续检测
        public NodeType Type = NodeType.Normal;	
        //……
    
        public void Divide(Collider collider)
        {
            //因为是正方体,所以用一条边来判断尺寸即可
            if(NodeCube.size.x >= MIN_CUBE_SIZE) 
            {
                //……
            }
            else
            {
                Type = NodeType.Obstacles;
            }
        }
        //isSeeOne为true,则只查看分裂后的一个,否则查看所有分裂后的方块
        public void Draw(bool isSeeOne)
        {
            var drawColor = Color.green;
            if(Type == NodeType.Obstacles)
                drawColor = Color.red;
            Gizmos.color = drawColor;
            //……
        }
    }
    

    可以清楚的看到排除的结点:

  2. 显然,我们需要用到图结构。由于本文的重点是在八叉树上,所以就不赘述图的实现了,作为一种基础的数据结构,我希望你能够自己实现。当然,实在没有的话,这里也提供一份作为参考吧(⊙ˍ⊙):

    using System.Collections.Generic;
    
    public class MyGraph<TNode, TEdge>
    {
    	public readonly HashSet<TNode> NodeSet;//结点列表
    	public readonly Dictionary<TNode, List<TNode>> NeighborList;//邻居列表
    	public readonly Dictionary<(TNode, TNode), List<TEdge>> EdgeList;//边列表
    	
        public MyGraph()
    	{
    		NodeSet = new HashSet<TNode>();
    		NeighborList = new Dictionary<TNode, List<TNode>>();
    		EdgeList = new Dictionary<(TNode, TNode), List<TEdge>>();
        }
    
    	/// <summary>
    	/// 寻找指定结点
    	/// </summary>
    	/// <returns>找到的结点,没找到时返回null</returns>
    	public TNode FindNode(TNode node)
    	{
    		NodeSet.TryGetValue(node, out TNode res);
    		return res;
    	}
    
    	/// <summary>
    	/// 寻找指点起、终点之间直接连接的所有边
    	/// </summary>
    	/// <param name="source">起点</param>
    	/// <param name="target">终点</param>
    	/// <returns>找到的边,没找到时返回null</returns>
    	public List<TEdge> FindEdge(TNode source, TNode target)
    	{
    		var s = FindNode(source);
    		var t = FindNode(target);
    		if (s != null && t != null)
    		{
    			var nodePairs = (s, t);
    			if (!EdgeList.ContainsKey(nodePairs))
    			{
    				return EdgeList[nodePairs];
    			}
    		}
    		return null;
    	}
    
    	/// <summary>
    	/// 添加结点,用HashSet,包含重复检测
    	/// </summary>
    	public bool AddNode(TNode node)
    	{
    		return NodeSet.Add(node);
    	}
    
    	/// <summary>
    	/// 添加指定边,含空结点判断、重复添加判断
    	/// </summary>
    	/// <param name="source">边起点</param>
    	/// <param name="target">边终点</param>
    	/// <param name="edge">指定边</param>
    	/// <returns>添加成功与否</returns>
    	public bool AddEdge(TNode source, TNode target, TEdge edge)
    	{
    		var s = FindNode(source);
    		var t = FindNode(target);
    		if (s == null || t == null)
    			return false;
    		var nodePairs = (s, t);
    		if(!EdgeList.ContainsKey(nodePairs))
    		{
    			EdgeList.Add(nodePairs, new List<TEdge>());
    		}
    		var allEdges = EdgeList[nodePairs];
    		if(!allEdges.Contains(edge))
    		{
    			allEdges.Add(edge);
    			if(!NeighborList.ContainsKey(source))
    			{
    				NeighborList.Add(source, new List<TNode>());
    			}
    			NeighborList[source].Add(target);
    			return true;
    		}
    		return false;
        }
    
    	/// <summary>
    	/// 移除指定结点
    	/// </summary>
    	/// <returns>移除成功与否</returns>
    	public bool RemoveNode(TNode node)
    	{
    		return NodeSet.Remove(node);
    	}
    
    	/// <summary>
    	/// 移除指定起、终点的指定边
    	/// </summary>
    	/// <param name="source">边起点</param>
    	/// <param name="target">边终点</param>
    	/// <param name="edge">指定边</param>
    	/// <returns>移除成功与否</returns>
    	public bool RemoveEdge(TNode source, TNode target, TEdge edge)
    	{
    		var allEdges = FindEdge(source, target);
    		return allEdges != null && allEdges.Remove(edge);
    	}
    
    	/// <summary>
    	/// 移除指定起、终点的所有边
    	/// </summary>
    	/// <param name="source">边起点</param>
    	/// <param name="target">边终点</param>
    	/// <returns>移除成功与否</returns>
    	public bool RemoveEdgeList(TNode source, TNode target)
    	{
    		return EdgeList.Remove((source, target));
    	}
    
    	/// <summary>
    	/// 获取指定结点可抵达的所有邻居结点
    	/// </summary>
    	public List<TNode> GetNeighbor(TNode node)
    	{
    		return NeighborList[node];
    	}
    
    	/// <summary>
    	/// 获取指定结点所延伸出的所有边
    	/// </summary>
    	public List<TEdge> GetConnectedEdge(TNode node)
    	{
    		var resEdge = new List<TEdge>();
    		var neighbor = GetNeighbor(node);
    		for(int i = 0; i < neighbor.Count; ++i)
    		{
    			var curEdgeList = EdgeList[(node, neighbor[i])];
    			for(int j = 0; j < curEdgeList.Count; ++j)
    			{
    				resEdge.Add(curEdgeList[j]);
    			}
    		}
    		return resEdge;
    	}
    }
    

    接下来就是让结点入图,我们在MyOctree类中声明一个图,并将树中所有正常结点都传入图中,这里也修改下构造函数,让图从外部传入(因为最终我们想要只操作MyOctreeBuilder脚本就能实现八叉树构建,所以把这些工作留给MyOctreeBuilder):

    public class MyOctree
    {
    	public MyOctreeNode RootNode;
    	public MyGraph<MyOctreeNode, int> NavGraph; //寻路网格
    	public MyOctree(GameObject[] allObjects)
    	{
    		var baseCube = new Bounds();
    		NavGraph = new MyGraph<MyOctreeNode, int>();
    
            //……
    
    		NodeToGraph(RootNode);
    	}
    	
    	//将树中的所有有效结点入图
    	private void NodeToGraph(MyOctreeNode node)
    	{
    		if (node == null) return;
    		// 没有子节点且为非障碍的结点才能入图
    		if(node.Children == null && node.Type != NodeType.Obstacles)
    		{
    			NavGraph.AddNode(node);
    		}
    		if(node.Children != null)
    		{
    			foreach(var c in node.Children)
    			{
    				NodeToGraph(c);
    			}
    		}
    	}
    }
    
    • (node.Children == null && node.Type != NodeType.Obstacles) 条件能剔除所有障碍结点?
      是可以做的,我们来看看下面几种情况:
      1. 有子节点的障碍方块。以下图绿色十字星的 \(4\times4\) 方形为例,它会被剔除
      2. 无子节点的障碍方块。仍是以绿色十字星标记的方形为例,它也会被剔除

    大改一下MyOctreeBuilder的内容,让它能绘制图也能绘制树,并根据功能开关绘制的内容:

    public class MyOctreeBuilder : MonoBehaviour
    {
    	public GameObject[] Objects; //场景包含的全部对象
    	public MyOctree Octree; // 八叉树
    	[SerializeField] private bool isSeeOne = false; //是否只观察一个分裂后的节点
    	[SerializeField] private bool isDrawOctreeCube = true; //是否绘制二叉树
    	[SerializeField] private bool isDrawNode = true; // 是否要绘制图的节点
    	[SerializeField] private bool isDrawEdge = true; // 是否要绘制图的边
    
    	private void Awake()
    	{
    		Octree = new MyOctree(Objects, new MyGraph<MyOctreeNode, int>());
    	}
    
    	private void OnDrawGizmos()
    	{
    		if(Application.isPlaying)
    		{
    			if(isDrawOctreeCube)
    			{
    				Octree.RootNode.Draw(isSeeOne);
    			}
    			DrawGraph();
    		}
    	}
    	
        private void DrawGraph()
    	{
    		if(isDrawEdge)
    		{
    			foreach(var edge in Octree.NavGraph.EdgeList)
    			{
    				Gizmos.color = Color.red;
    				Gizmos.DrawLine(edge.Key.Item1.NodeCube.center, 
    				edge.Key.Item2.NodeCube.center);
    			}				
    		}
    		if(isDrawNode)
    		{
    			foreach(var node in Octree.NavGraph.NodeSet)
    			{
    				Gizmos.color = new Color(1, 1, 0);
    				Gizmos.DrawWireSphere(node.NodeCube.center, 0.25f);
    			}				
    		}
    	}
    }
    

    可以看到,障碍物内部是没有结点的,障碍结点都被剔除了:

    最后,就将这些结点连接起来吧:

    public class MyOctree
    {
    	public MyOctreeNode RootNode;
    	public MyGraph<MyOctreeNode, int> NavGraph; //寻路网格图
    	public MyOctree(GameObject[] allObjects)
    	{
            //……
    
    		NodeToGraph(RootNode);
    		GenerateEdges();
    	}
    	
        //……
    
    	//生成边
    	private void GenerateEdges()
    	{
    		foreach(var f in NavGraph.NodeSet)
    		{
    			foreach(var t in NavGraph.NodeSet)
    			{
    				if (f == t)
    					continue;
    				var ray = new Ray(f.NodeCube.center, t.NodeCube.center - f.NodeCube.center);
    				// 限制全连接范围
    				var maxDistance = f.NodeCube.size.y * 0.7f;
    				if(t.NodeCube.IntersectRay(ray, out float hitDistance))
    				{
    					if (hitDistance > maxDistance)
    						continue;
    					// 添加无向边(双向),路径长度默认为1,如有需求可自行调整
    					NavGraph.AddEdge(f, t, 1);
    					NavGraph.AddEdge(t, f, 1);
    				}
    			}
    		}
    	} 
    }
    

    最后的样子(共大概600个结点、4000多条边):

立体网格寻路

网格已经构建完成,离寻路还差最后一步了。或许有的同学只知道在平面地图寻路的A*算法实现,怎么将它应用在立体地图中?

咳咳,不是打广告啊= ̄ω ̄=,也许这篇文章能对你有所帮助,那是我尝试的一个泛用A*搜索的模板,简单实现相关接口就可以在这种地图进行A*寻路了:

PS:个人与2024-6-1优化了上述文章中的优先队列(堆)的实现,所以整体代码有了小变动,如果你是在2024-6-1之后看的,那请忽视这句话。

public class MyOctreeNode:IAStarNode<MyOctreeNode>, IComparable<MyOctreeNode>
{
    private const float MIN_CUBE_SIZE = 1f; // 最小方格尺寸
    public MyOctreeNode Parent{ get; set; } //父节点
    
    //实现IAStarNode接口属性
    public float SelfCost { get; set; }
    public float GCost { get; set; }
    public float HCost { get; set; }
    public float FCost => GCost + HCost;

    //……

    //实现接口函数
    public float GetDistance(MyOctreeNode otherNode)
    {
        return Vector3.Distance(NodeCube.center, otherNode.NodeCube.center);
    }

    public List<MyOctreeNode> GetSuccessors(object nodeMap)
    {
        var map = (MyGraph<MyOctreeNode, int>)nodeMap;
        return map.GetNeighbor(this);
    }

    public int CompareTo(MyOctreeNode other)
    {
        float res = FCost - other.FCost;
        if(res == 0)
            res = HCost - other.HCost;
        return (int)res;
    }
}

还有一件事,要实现一个将空间点转化为八叉树节点的方法,这也不难,就是可以通过Bounds.Contains方法查询一个点是否在包围盒内部,我们在MyOctree类中添加这样的方法:

/// <summary>
/// 以指定节点开始搜索,寻找到与指定位置最接近的节点
/// </summary>
/// <param name="start">初始点</param>
/// <param name="pos">指定位置</param>
/// <returns>寻找到的节点,若没找到则返回根节点</returns>
public MyOctreeNode GetNodeByPos(MyOctreeNode start, Vector3 pos)
{
    MyOctreeNode findNode = RootNode;
    if (start == null) 
        return findNode;
    if (start.Children == null)
    {
        if(start.NodeCube.Contains(pos))
            return start;
    }
    else
    {
        for(int i = 0; i < 8; ++i)
        {
            findNode = GetNodeByPos(start.Children[i], pos);
            if (findNode != RootNode)
                return findNode;
        }				
    }
    return findNode;
}

最后,创建一个用来驱动A*搜索器的脚本MyOctreeAStar:

using System.Collections;
using JufGame.AI;
using System.Collections.Generic;
using UnityEngine;

public class MyOctreeAStar : MonoBehaviour
{
    public MyOctreeBuilder octree; //八叉树构建器
    private AStar_Searcher<MyGraph<MyOctreeNode, int>, MyOctreeNode> astar; //A星搜索器
    private Stack<MyOctreeNode> path; //存储路径的栈
    [SerializeField] private Transform start; //寻路起点
    [SerializeField] private Transform end; //寻路终点
    //当该值位false时会进行一次寻路,寻路完成后自动为true
    [SerializeField] private bool isFindPathEnd; 
    private void Start()
    {
        astar = new AStar_Searcher<MyGraph<MyOctreeNode, int>, MyOctreeNode>(octree.Octree.NavGraph);
        path = new Stack<MyOctreeNode>();
    }
    private void Update()
    {
        if(!isFindPathEnd)
        {
            //将起点与终点的位置转化为树中节点,然后进行寻路
            var s = octree.Octree.GetNodeByPos(octree.Octree.RootNode, start.position);
            var e = octree.Octree.GetNodeByPos(octree.Octree.RootNode, end.position);
            astar.FindPath(s, e, path);
            isFindPathEnd = true;
        }
    }
    private void OnDrawGizmos()
    {
        if(Application.isPlaying)
        {
            var prevPos = start.position;
            foreach(var n in path)
            {
                Gizmos.color = Color.red;
                Gizmos.DrawLine(prevPos, n.NodeCube.center);
                prevPos = n.NodeCube.center;
            }
            Gizmos.DrawLine(prevPos, end.position);
        }
    }
}

将它挂在场景的一个物体中,设置好起点和终点(要保证起点和终点在八叉树覆盖的范围内,否则寻路会报错),然后就可以尝试寻路了:

结尾(完整代码)

利用八叉树的寻路基本就讲完了,在编写期间因为时不时对代码进行调整,可能导致各文段代码有前后差异(本人努力排查过几次了,但可能难免有疏漏),现贴上最终代码:

八叉树相关的四个类:

using System;
using System.Collections;
using System.Collections.Generic;
using JufGame.AI;
using UnityEngine;

public enum NodeType
{
    Normal, Obstacles,
}
public class MyOctreeNode:IAStarNode<MyOctreeNode>, IComparable<MyOctreeNode>
{
    private const float MIN_CUBE_SIZE = 1f; // 最小方格尺寸
    public MyOctreeNode Parent{ get; set; } //父节点
    public float SelfCost { get; set; }
    public float GCost { get; set; }
    public float HCost { get; set; }
    public float FCost => GCost + HCost;

    public MyOctreeNode[] Children; //子节点
    public Bounds NodeCube; //用包围盒作为结点方块,方便后续检测
    public NodeType Type = NodeType.Normal;
    public MyOctreeNode(Bounds nodeCube, MyOctreeNode parent)
    {
        Parent = parent;
        NodeCube = nodeCube;
        SelfCost = 1;
    }
    public void Divide(Collider collider)
    {
        //因为是正方体,所以用一条边来判断尺寸即可
        if(NodeCube.size.x >= MIN_CUBE_SIZE) 
        {
            // 子方块的半尺寸, 用半尺寸是因为构建Bounds需要
            float childHalfSize = NodeCube.size.x / 4;
            if (Children == null)
                Children = new MyOctreeNode[8];
            Vector3 offset; //子节点偏移
            for(int i = 0; i < 8; ++i)
            {
                //0~7的二进制位结构恰好满足我们所需要的组合形式
                offset.x = (1 & i) != 0 ? childHalfSize : -childHalfSize; //取二进制第0位
                offset.y = (2 & i) != 0 ? childHalfSize : -childHalfSize; //取二进制第1位
                offset.z = (4 & i) != 0 ? childHalfSize : -childHalfSize; //取二进制第2位
                var childBounds = new Bounds(NodeCube.center + offset, 2 * childHalfSize * Vector3.one);
                if(Children[i] == null)
                    Children[i] = new MyOctreeNode(childBounds, this);
                /*
                进一步分裂前,先判断一下有没有遇到障碍物,没有就不要继续分裂了;
                也可以再附带添加些其它检测条件,比如obj.layer等
                */
                if(childBounds.Intersects(collider.bounds))
                {
                    Children[i].Divide(collider); // 每个子节点继续分裂
                }
            }
        }
        else
        {
            Type = NodeType.Obstacles;
        }
    }
    //seeOne为true,则只查看分裂后的一个,否则查看所有分裂后的方块
    public void Draw(bool isSeeOne)
    {
        var drawColor = Color.green;
        if(Type == NodeType.Obstacles)
            drawColor = Color.red;
        Gizmos.color = drawColor;
        Gizmos.DrawWireCube(NodeCube.center, NodeCube.size);
        if (Children == null)
            return;
        foreach(var c in Children)
        {
            c.Draw(isSeeOne);
            if(isSeeOne)
            {
                break;
            }
        }
    }

    public float GetDistance(MyOctreeNode otherNode)
    {
        return Vector3.Distance(NodeCube.center, otherNode.NodeCube.center);
    }

    public List<MyOctreeNode> GetSuccessors(object nodeMap)
    {
        var map = (MyGraph<MyOctreeNode, int>)nodeMap;
        return map.GetNeighbor(this);
    }

    public int CompareTo(MyOctreeNode other)
    {
        float res = FCost - other.FCost;
        if(res == 0)
            res = HCost - other.HCost;
        return (int)res;
    }
}
using System.Collections;
using System.Collections.Generic;
using UnityEngine;

public class MyOctree
{
    public MyOctreeNode RootNode;
    public MyGraph<MyOctreeNode, int> NavGraph; //寻路网格图
    public MyOctree(GameObject[] allObjects, MyGraph<MyOctreeNode, int> navGraph)
    {
        var baseCube = new Bounds();
        NavGraph = navGraph;
        foreach(var o in allObjects)
        {
            baseCube.Encapsulate(o.GetComponent<Collider>().bounds);
        }
        //选取最长的一条边来作为正方体的边长,并将包围盒改成正方体
        //这里为了更好设置包围盒,同样记录半尺寸
        var cubeHalfSize = 0.5f * Mathf.Max(baseCube.size.x, baseCube.size.y, baseCube.size.z) * Vector3.one;
        baseCube.SetMinMax(baseCube.center - cubeHalfSize, baseCube.center + cubeHalfSize);

        RootNode = new MyOctreeNode(baseCube, null);
        foreach(var o in allObjects) //具体分裂
        {
            //有碰撞体的物体才有检测的必要
            if(o.TryGetComponent(out Collider collider))
            {
                RootNode.Divide(collider);
            }	
        }
        NodeToGraph(RootNode);
        //Debug.Log(NavGraph.NodeSet.Count); //查看结点数量
        GenerateEdges();
        //Debug.Log(NavGraph.EdgeList.Count); //查看边的数量
    }
    
    //将树中的所有结点入图
    private void NodeToGraph(MyOctreeNode node)
    {
        if (node == null) return;
        // 没有子节点且为非障碍的结点才能入图
        if(node.Children == null && node.Type != NodeType.Obstacles)
        {
            NavGraph.AddNode(node);
        }
        if(node.Children != null)
        {
            foreach(var c in node.Children)
            {
                NodeToGraph(c);
            }
        }
    }

    //生成边
    private void GenerateEdges()
    {
        foreach(var f in NavGraph.NodeSet)
        {
            foreach(var t in NavGraph.NodeSet)
            {
                if (f == t)
                    continue;
                var ray = new Ray(f.NodeCube.center, t.NodeCube.center - f.NodeCube.center);
                // 限制全连接范围
                var maxDistance = f.NodeCube.size.y * 0.7f;
                if(t.NodeCube.IntersectRay(ray, out float hitDistance))
                {
                    if (hitDistance > maxDistance)
                        continue;
                    // 添加无向边(双向),路径长度默认为1,如有需求可自行调整
                    NavGraph.AddEdge(f, t, 1);
                    NavGraph.AddEdge(t, f, 1);
                }
            }
        }
    } 

    /// <summary>
    /// 以指定节点开始搜索,寻找到与指定位置最接近的节点
    /// </summary>
    /// <param name="start">初始点</param>
    /// <param name="pos">指定位置</param>
    /// <returns>寻找到的节点,若没找到则返回根节点</returns>
    public MyOctreeNode GetNodeByPos(MyOctreeNode start, Vector3 pos)
    {
        MyOctreeNode findNode = RootNode;
        if (start == null) 
            return findNode;
        if (start.Children == null)
        {
            if(start.NodeCube.Contains(pos))
                return start;
        }
        else
        {
            for(int i = 0; i < 8; ++i)
            {
                findNode = GetNodeByPos(start.Children[i], pos);
                if (findNode != RootNode)
                    return findNode;
            }				
        }
        return findNode;
    }
}
using System.Collections;
using System.Collections.Generic;
using UnityEngine;

public class MyOctreeBuilder : MonoBehaviour
{
    public GameObject[] Objects; //场景包含的全部对象
    public MyOctree Octree; // 八叉树
    [SerializeField] private bool isSeeOne = false; //是否只观察一个分裂后的节点
    [SerializeField] private bool isDrawOctreeCube = true; //是否绘制二叉树
    [SerializeField] private bool isDrawNode = true; // 是否要绘制图的节点
    [SerializeField] private bool isDrawEdge = true; // 是否要绘制图的边
    private void Awake()
    {
        Octree = new MyOctree(Objects, new MyGraph<MyOctreeNode, int>());
    }
    private void OnDrawGizmos()
    {
        if(Application.isPlaying)
        {
            if(isDrawOctreeCube)
            {
                Octree.RootNode.Draw(isSeeOne);
            }
            DrawGraph();
        }
    }
    private void DrawGraph()
    {
        if(isDrawEdge)
        {
            foreach(var edge in Octree.NavGraph.EdgeList)
            {
                Gizmos.color = Color.red;
                Gizmos.DrawLine(edge.Key.Item1.NodeCube.center, 
                edge.Key.Item2.NodeCube.center);
            }				
        }
        if(isDrawNode)
        {
            foreach(var node in Octree.NavGraph.NodeSet)
            {
                Gizmos.color = new Color(1, 1, 0);
                Gizmos.DrawWireSphere(node.NodeCube.center, 0.25f);
            }				
        }
    }
}
using System.Collections;
using JufGame.AI;
using System.Collections.Generic;
using UnityEngine;

public class MyOctreeAStar : MonoBehaviour
{
    public MyOctreeBuilder octree; //八叉树构建器
    private AStar_Searcher<MyGraph<MyOctreeNode, int>, MyOctreeNode> astar; //A星搜索器
    private Stack<MyOctreeNode> path; //存储路径的栈
    [SerializeField] private Transform start; //寻路起点
    [SerializeField] private Transform end; //寻路终点
    //当该值位false时会进行一次寻路,寻路完成后自动为true
    [SerializeField] private bool isFindPathEnd; 
    private void Start()
    {
        astar = new AStar_Searcher<MyGraph<MyOctreeNode, int>, MyOctreeNode>(octree.Octree.NavGraph);
        path = new Stack<MyOctreeNode>();
    }
    private void Update()
    {
        if(!isFindPathEnd)
        {
            //将起点与终点的位置转化为树中节点,然后进行寻路
            var s = octree.Octree.GetNodeByPos(octree.Octree.RootNode, start.position);
            var e = octree.Octree.GetNodeByPos(octree.Octree.RootNode, end.position);
            astar.FindPath(s, e, path);
            isFindPathEnd = true;
        }
    }
    private void OnDrawGizmos()
    {
        if(Application.isPlaying)
        {
            var prevPos = start.position;
            foreach(var n in path)
            {
                Gizmos.color = Color.red;
                Gizmos.DrawLine(prevPos, n.NodeCube.center);
                prevPos = n.NodeCube.center;
            }
            Gizmos.DrawLine(prevPos, end.position);
        }
    }
}

与A星相关的代码也贴这里了:

using System;
using System.Collections.Generic;

namespace JufGame.Collections.Generic
{
    public class MyHeap<T> where T : IComparable<T>
    {
        public int NowLength { get; private set; }
        public int MaxLength { get; private set; }
        public T Top => heap[0];
        public bool IsEmpty => NowLength == 0;
        public bool IsFull => NowLength >= MaxLength - 1;
        private readonly Dictionary<T, int> nodeIdxTable; // 记录结点在数组中的位置,方便查找
        private readonly bool isReverse;
        private readonly T[] heap;

        public MyHeap(int maxLength, bool isReverse = false)
        {
            NowLength = 0;
            MaxLength = maxLength;
            heap = new T[MaxLength + 1];
            nodeIdxTable = new Dictionary<T, int>();
            this.isReverse = isReverse;
        }
        public T this[int index]
        {
            get => heap[index];
        }
        public void PushHeap(T value)
        {
            if (NowLength < MaxLength)
            {
                if (nodeIdxTable.ContainsKey(value))
                    nodeIdxTable[value] = NowLength;
                else
                    nodeIdxTable.Add(value, NowLength);
                heap[NowLength] = value;
                Swim(NowLength);
                ++NowLength;
            }
        }
        public void PopHeap()
        {
            if (NowLength > 0)
            {
                nodeIdxTable[heap[0]] = -1; 
                heap[0] = heap[--NowLength];
                nodeIdxTable[heap[0]] = 0;
                Sink(0);
            }
        }
        public bool Contains(T value)
        {
            return nodeIdxTable.ContainsKey(value) && nodeIdxTable[value] != -1;
        }
        public T Find(T value)
        {
            if (Contains(value))
                return heap[nodeIdxTable[value]];
            return default;
        }
        public void Clear()
        {
            nodeIdxTable.Clear();
            NowLength = 0;
        }
        private void SwapValue(T a, T b)
        {
            var aIdx = nodeIdxTable[a];
            var bIdx = nodeIdxTable[b];
            heap[aIdx] = b;
            heap[bIdx] = a;
            nodeIdxTable[a] = bIdx;
            nodeIdxTable[b] = aIdx;
        }

        private void Swim(int index)
        {
            int father;
            while (index > 0)
            {
                father = (index - 1) >> 1;
                if (IsBetter(heap[index], heap[father]))
                {
                    SwapValue(heap[father], heap[index]);
                    index = father;
                }
                else return;
            }
        }

        private void Sink(int index)
        {
            int largest, left = (index << 1) + 1;
            while (left < NowLength)
            {
                largest = left + 1 < NowLength && IsBetter(heap[left + 1], heap[left]) ? left + 1 : left;
                if (IsBetter(heap[index], heap[largest]))
                    largest = index;
                if (largest == index) return;
                SwapValue(heap[largest], heap[index]);
                index = largest;
                left = (index << 1) + 1;
            }
        }
        private bool IsBetter(T v1, T v2)
        {
            return isReverse ? (v2.CompareTo(v1) < 0 ): (v1.CompareTo(v2) < 0);
        }
    }
}
using JufGame.Collections.Generic;
using System;
using System.Collections.Generic;

namespace JufGame.AI
{
	public interface IAStarNode<T> where T : IAStarNode<T>
	{
        public T Parent { get; set; }
        public float SelfCost { get; set; }
        public float GCost { get; set; }//距初始状态的代价
        public float HCost { get; set; }//距目标状态的代价
        public float FCost { get; }
        /// <summary>
        /// 获取与指定节点的预测代价
        /// </summary>
        public float GetDistance(T otherNode);
		/// <summary>
		/// 获取后继(邻居)节点
		/// </summary>
        /// <param name="nodeMap">寻路所在的地图,类型看具体情况转换,
        /// 故用object类型</param>
        /// <returns>后继节点列表</returns>
		public List<T> GetSuccessors(object nodeMap);
        /* 一般比较可用以下函数
        public int CompareTo(AStarNode other)
        {
        	var res = (int)(FCost - other.FCost);
			if(res == 0)
				res = (int)(HCost - other.HCost);
			return res;
        }
        */
	}
    /// <summary>
    /// A星搜索器
    /// </summary>
    /// <typeparam name="T_Map">搜索的图类</typeparam>
    /// <typeparam name="T_Node">搜索的节点类</typeparam>
	public class AStar_Searcher<T_Map, T_Node> where T_Node: IAStarNode<T_Node>, IComparable<T_Node>
	{
        private readonly HashSet<T_Node> closeList;//探索集
        private readonly MyHeap<T_Node> openList;//边缘集
        private readonly T_Map nodeMap;//搜索空间(地图)
        public AStar_Searcher(T_Map map, int maxNodeSize = 200)
        {
            nodeMap = map;
            closeList = new HashSet<T_Node>();
            //maxNodeSize用于限制路径节点的上限,避免陷入无止境搜索的情况
            openList = new MyHeap<T_Node>(maxNodeSize);
        }
        /// <summary>
        /// 搜索(寻路)
        /// </summary>
        /// <param name="start">起点</param>
        /// <param name="target">终点</param>
        /// <param name="pathRes">返回生成的路径</param>
        public void FindPath(T_Node start, T_Node target, Stack<T_Node> pathRes)
        {
            T_Node currentNode;
            pathRes.Clear();//清空路径以备存储新的路径
            closeList.Clear();
            openList.Clear();
            openList.PushHeap(start);
            while (!openList.IsEmpty)
            {
                currentNode = openList.Top;//取出边缘集中最小代价的节点
                openList.PopHeap();
                closeList.Add(currentNode);//拟定移动到该节点,将其放入探索集
                if (currentNode.Equals(target) || openList.IsFull)//如果找到了或图都搜完了也没找到时
                {
                    GenerateFinalPath(start, target, pathRes);//生成路径并保存到pathRes中
                    return;
                }
                UpdateList(currentNode, target);//更新边缘集和探索集
            }
            return;
        }
        private void GenerateFinalPath(T_Node startNode, T_Node endNode, Stack<T_Node> pathStack)
        {
            pathStack.Push(endNode);//因为回溯,所以用栈储存生成的路径
            var tpNode = endNode.Parent;
            while (!tpNode.Equals(startNode))
            {
                pathStack.Push(tpNode);
                tpNode = tpNode.Parent;
            }
            pathStack.Push(startNode);
        }
        private void UpdateList(T_Node curNode, T_Node endNode)
        {
            T_Node sucNode;
            float tpCost;
            bool isNotInOpenList;
            var successors = curNode.GetSuccessors(nodeMap);//找出当前节点的后继节点
            for (int i = 0; i < successors.Count; ++i)
            {
                sucNode = successors[i];
                if (closeList.Contains(sucNode))//后继节点已被探索过就忽略
                    continue;
                tpCost = curNode.GCost + sucNode.SelfCost;
                isNotInOpenList = !openList.Contains(sucNode);
                if (isNotInOpenList || tpCost < sucNode.GCost)
                {
                    sucNode.GCost = tpCost;
                    sucNode.HCost = sucNode.GetDistance(endNode);//计算启发函数估计值
                    sucNode.Parent = curNode;//记录父节点,方便回溯
                    if (isNotInOpenList)
                    {
                        openList.PushHeap(sucNode);
                    }
                }
            }
        }
    }
}

在尝试用渐进式的方式写文章,基本就是顺着思路走下来的 (如果有某几步跳得比较大,那就是我写烦了。所以文中的代码修改会比较频繁,但我觉得比起先将思路再贴出完整代码的方式,这样会更容易让人理解(当然,只是个人觉得。

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