PAT L3-008. 喊山(BFS)C4 初赛30分】的更多相关文章

喊山(30 分) 喊山,是人双手围在嘴边成喇叭状,对着远方高山发出“喂—喂喂—喂喂喂……”的呼唤.呼唤声通过空气的传递,回荡于深谷之间,传送到人们耳中,发出约定俗成的“讯号”,达到声讯传递交流的目的.原来它是彝族先民用来求援呼救的“讯号”,慢慢地人们在生活实践中发现了它的实用价值,便把它作为一种交流工具世代传袭使用.(图文摘自:http://news.xrxxw.com/newsshow-8018.html) 一个山头呼喊的声音可以被临近的山头同时听到.题目假设每个山头最多有两个能听到它的临近山…
PAT甲级:1064 Complete Binary Search Tree (30分) 题干 A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties: The left subtree of a node contains only nodes with keys less than the node's key. The right subtre…
L3-008. 喊山 时间限制 150 ms 内存限制 65536 kB 代码长度限制 8000 B 判题程序 Standard 作者 陈越 喊山,是人双手围在嘴边成喇叭状,对着远方高山发出“喂—喂喂—喂喂喂……”的呼唤.呼唤声通过空气的传递,回荡于深谷之间,传送到人们耳中,发出约定俗成的“讯号”,达到声讯传递交流的目的.原来它是彝族先民用来求援呼救的“讯号”,慢慢地人们在生活实践中发现了它的实用价值,便把它作为一种交流工具世代传袭使用.(图文摘自:http://news.xrxxw.com/n…
一个山头呼喊的声音可以被临近的山头同时听到.题目假设每个山头最多有两个能听到它的临近山头.给定任意一个发出原始信号的山头,本题请你找出这个信号最远能传达到的地方. 输入格式: 输入第一行给出3个正整数n.m和k,其中n(≤10000)是总的山头数(于是假设每个山头从1到n编号).接下来的m行,每行给出2个不超过n的正整数,数字间用空格分开,分别代表可以听到彼此的两个山头的编号.这里保证每一对山头只被输入一次,不会有重复的关系输入.最后一行给出k(≤10)个不超过n的正整数,数字间用空格分开,代表…
A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties: The left subtree of a node contains only nodes with keys less than the node's key. The right subtree of a node contains only nodes with keys greate…
A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties: The left subtree of a node contains only nodes with keys less than or equal to the node's key. The right subtree of a node contains only nodes with…
The lowest common ancestor (LCA) of two nodes U and V in a tree is the deepest node that has both U and V as descendants. Given any two nodes in a binary tree, you are supposed to find their LCA. Input Specification: Each input file contains one test…
1053 Path of Equal Weight(30 分) Given a non-empty tree with root R, and with weight W​i​​ assigned to each tree node T​i​​. The weight of a path from R to L is defined to be the sum of the weights of all the nodes along the path from R to any leaf no…
将一系列给定数字顺序插入一个初始为空的二叉搜索树(定义为左子树键值大,右子树键值小),你需要判断最后的树是否一棵完全二叉树,并且给出其层序遍历的结果. 输入格式: 输入第一行给出一个不超过20的正整数N:第二行给出N个互不相同的正整数,其间以空格分隔. 输出格式: 将输入的N个正整数顺序插入一个初始为空的二叉搜索树.在第一行中输出结果树的层序遍历结果,数字间以1个空格分隔,行的首尾不得有多余空格.第二行输出YES,如果该树是完全二叉树:否则输出NO. 输入样例1: 9 38 45 42 24 5…
题意:输入一个正整数N(<=1000),接着输入N个非负整数(<=2000),输出完全二叉树的层次遍历. AAAAAccepted code: #define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std; ]; ]; int cnt; int n; void dfs(int x){ if(x>n) return; dfs(x*); tree[x]=a[++cnt]; dfs(x*+); }…