aabb):ifself.rootisNone:self.root=AABBTreeNode(aabb)else:self._insert_rec(self.root,aabb)def_insert_rec(self,node,aabb):ifnode.leftisNoneandnode.rightisNone:node.left=AABBTreeNode(aabb)elifnode.leftisnotNoneandnode.rightisnotNone:# Find a node to insert intoifnode.left.aabb.intersects...
if __name__ == "__main__": tree = Tree() tree.list_add([0,1,2,3,4,5,6,7,8,9]) tree.add(10) ls=tree.breadth_travel() print(ls) ls = tree.preorder() print(ls) ls=tree.inorder() print(ls) ls=tree.postorder() print(ls) 3结语 ...
FCL还支持对象组之间的Broadphase 碰撞/距离查询,并且可以避免成对计算带来的n^2的复杂度。 具体地,在执行 碰撞/距离检查 之前,将CollisionObject项注册到DynamicAABBTreeCollisionManager中。 提供三种类型的检查: 一对多: 单独的CollisionObject与Manager中所有对象之间的碰撞/距离检查。 Manager内的多对多: Manager中所...
创建python文件读取xmltest.xml内容 import xml.etree.ElementTree as ET tree = ET.parse("xmltest.xml") # 写上要处理的文件明智 root = tree.getroot() print(root) # 打印的是内存地址信息 print(root.tag) # root.tag = data其实就是标签的名字 # 遍历xml文档 for child in root: # 循环每个子标...
5 importnumpy as np importpandas as pd importmatplotlib importmatplotlib.pyplot as plt importseaborn as sns matplotlib参数设置 1 2 3 4 5 6 7 8 9 10 11 matplotlib.rcParams['font.sans-serif']=['SimHei'] matplotlib.rcParams['font.family']='sans-serif' ...
row=2)self.wifi_mm_input=Entry(labelframe,width=10,textvariable=self.get_wifimm_value).grid(column=3,row=2,sticky=W)self.wifi_labelframe=LabelFrame(text="wifi列表")self.wifi_labelframe.grid(column=0,row=3,columnspan=4,sticky=NSEW)# 定义树形结构与滚动条self.wifi_tree=ttk.Treeview(self...
4 Explanation: Replace the two 'A's with two 'B's or vice versa. Example 2: Input: s = "AABABBA", k = 1 Output: 4 Explanation: Replace the one 'A' in the middle with 'B' and form "AABBBBA". The substring "BBBB" has the longest repeating letters, which is 4. ...
如果是3D的话,可以使用八叉树和BVH。BVH的查询是比八叉树快,但是在考虑到动态更新的情况下八叉树更好,BVH需要更新当前物体所在节点的整个分支,而八叉树只需要更新旧位置和新位置的两个节点。(BVH的动态更新参考Game Physics: Broadphase – Dynamic AABB Tree)...
利用KD-Tree的结构来构建AABB的好处是倘若光线与哪一部分空间不相交,那么则可以省略该部分空间所有子空间的判断过程,在原始的纯粹的AABB之上更进一步提升了加速效率。 缺点: 缺点是判断包围盒与三角面的是否相交较难,因此划分的过程不是那么想象的简单,其次同一个三角面可能被不同的包围盒同时占有,这两个不同包围盒...
n = int(input()) height = 1 for i in range(n): if (i + 1) % 2 == 1: height *= 2 else: height += 1 print(height) 递归实现 num = eval(input()) def tree(n): if n == 0: return 1 elif n == 1: return 2 elif n%2: return 2*(tree(n-2) + 1) else: return ...