1021 Deepest Root (25 分)   A graph which is connected and acyclic can be considered a tree. The height of the tree depends on the selected root. Now you are supposed to find the root that results in a highest tree. Such a root is called the deepest r…

题目 A supply chain is a network of retailers(零售商), distributors(经销商), and suppliers(供应商)– everyone involved in moving a product from supplier to customer. Starting from one root supplier, everyone on the chain buys products from one's supplier in a pr…

题目 A supply chain is a network of retailers(零售商), distributors(经销商), and suppliers(供应商)– everyone involved in moving a product from supplier to customer. Starting from one root supplier, everyone on the chain buys products from one's supplier in a pr…

题目链接:https://pintia.cn/problem-sets/994805046380707840/problems/994805069361299456 L2-006 树的遍历 (25 分)   给定一棵二叉树的后序遍历和中序遍历,请你输出其层序遍历的序列.这里假设键值都是互不相等的正整数. 输入格式: 输入第一行给出一个正整数N(≤30),是二叉树中结点的个数.第二行给出其后序遍历序列.第三行给出其中序遍历序列.数字间以空格分隔. 输出格式: 在一行中输出该树的层序遍历的序列.数字…

1444 破坏道路 题目来源: CodeForces 基准时间限制:1.5 秒 空间限制:131072 KB 分值: 80 难度:5级算法题 收藏 关注 在某一个国家,那儿有n个城市,他们通过m条双向道路相连.城市从1到n编号.如果城市a和b通过一条道路直接相连,那么他们之间的距离就是一个小时.这个国家的道路网络可以允许你从任意一个城市到达另外的城市. 现在你要破坏尽可能多的道路,但是要保证从城市s1到t1不超过l1小时,并且从城市s2到t2不超过l2小时. 输出最多可以破坏的道路数目,如果没有…

Knight Moves A friend of you is doing research on the Traveling Knight Problem (TKP) where you are to find the shortest closed tour of knight moves that visits each square of a given set of n squares on a chessboard exactly once. He thinks that the m…

# 逻辑回归 ## 逻辑回归处理二元分类 %matplotlib inline import matplotlib.pyplot as plt #显示中文 from matplotlib.font_manager import FontProperties font=FontProperties(fname=r"c:\windows\fonts\msyh.ttc", size=10) import numpy as np plt.figure() plt.axis([-6,6,0,1]…

6.标签特征二元化 处理分类变量还有另一种方法,不需要通过OneHotEncoder,我们可以用LabelBinarizer. 这是一个阈值与分类变量组合的方法. In [1]: from sklearn import datasets as d iris = d.load_iris() target = iris.target How to do it... 导入LabelBinarizer()创建一个对象: In [2]: from sklearn.preprocessing import…

Python List方法总结 一. 列表简介: 列表是序列对象,可包含任意的Python数据信息,如字符串.数字.列表.元组等   列表的数据是可变的,我们可通过对象方法对列表中的数据进行增加.修改.删除等操作   可以通过list(seq)函数把一个序列类型转换成一个列表 运算符: 索引运算[i] ,切片运算[i:j], 扩展切片运算[i:j:stride] 支持运算:索引,切片,min(),max(),len()等 1. 列表赋值: l1=[] 空列表   #但是内存有位置存放,使用id(…

map #对参数迭代器中的每个元素进行操作,返回一个新的迭代器 map(func, *iterables) --> map object Make an iterator that computes the function using arguments from each of the iterables. Stops when the shortest iterable is exhausted. >>> l1=[1,3,5,7,9] #求列表l1中的每个元素的平方 >…