PAT甲级——A1021 Deepest Root

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 root. Input Specification: E…

PAT甲级1021. Deepest Root

PAT甲级1021. Deepest Root 题意: 连接和非循环的图可以被认为是一棵树.树的高度取决于所选的根.现在你应该找到导致最高树的根.这样的根称为最深根. 输入规格: 每个输入文件包含一个测试用例.对于每种情况, 第一行包含正整数N(<= 10000),它是节点的数量,因此节点从1到N编号.然后按N-1行,每个都通过给定两个相邻节点的数字来描述一个边. 输出规格: 对于每个测试用例,打印一行中最深的根.如果这样的根不是唯一的, 打印他们的数字增加的顺序.在给定的图形不是树的情况下,打…

PAT 甲级 1021 Deepest Root

https://pintia.cn/problem-sets/994805342720868352/problems/994805482919673856 A graph which is connected and acyclic can be considered a tree. The hight of the tree depends on the selected root. Now you are supposed to find the root that results in a…

PAT 甲级 1021 Deepest Root (并查集，树的遍历)

1021. Deepest Root (25) 时间限制 1500 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue 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 t…

PAT 甲级 1021 Deepest Root (25 分)（bfs求树高,又可能存在part数part>2的情况）

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…

PAT Advanced A1021 Deepest Root (25) [图的遍历，DFS，计算连通分量的个数，BFS，并查集]

题目 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 root. Input Specification…

PAT甲级：1066 Root of AVL Tree (25分)

PAT甲级:1066 Root of AVL Tree (25分) 题干 An AVL tree is a self-balancing binary search tree. In an AVL tree, the heights of the two child subtrees of any node differ by at most one; if at any time they differ by more than one, rebalancing is done to rest…

PAT A1021 Deepest Root （25 分）——图的BFS,DFS

A graph which is connected and acyclic can be considered a tree. The hight 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 root. Input Specification: Ea…

A1021. Deepest Root

[PAT] A1021 Deepest Root

[题目大意] 给出n个结点和n-1条边,问它们能否形成一棵n个结点的树,如果能,从中选出结点作为树根,使整棵树的高度最大.输出所有满足要求的可以作为树根的结点. [思路] 方法一:模拟. 1 连通.边数为n-1的图一定是一棵树.因此先判断连通图个数(用DFS遍历图,从而计算连通图个数),等于1则能形成一棵树. 2 遍历每一个节点,假设它为根节点构造一棵树. 方法二: 1 由于连通.边数为n-1的图一定是一棵树,因此需要判断给定数据是否能使图连通.2 确定图连通后,则确定了树,选择合适根结点使树高…

PAT甲级题解-1066. Root of AVL Tree (25)-AVL树模板题

博主欢迎转载,但请给出本文链接,我尊重你,你尊重我,谢谢~http://www.cnblogs.com/chenxiwenruo/p/6803291.html特别不喜欢那些随便转载别人的原创文章又不给出链接的所以不准偷偷复制博主的博客噢~~ 题意:给你一个插入的序列,问你最后AVL树的根节点是多少 AVL树模板题如果会AVL树,那么就是水题如果不会的话,那么就是难题我刚开始也不会,所以根本写不出来AVL树...后来花了一些时间学习了下,感觉网上很多模板都是用class写的,太麻烦了模板就是应该要…

【PAT甲级】1066 Root of AVL Tree (25 分)（AVL树建树模板）

题意: 输入一个正整数N(<=20),接着输入N个结点的值,依次插入一颗AVL树,输出最终根结点的值. AAAAAccepted code: #define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std; ]; typedef struct node{ int val; node*left,*right; }; node*left_rotate(node*root){ node*tamp=root->…

PAT_A1021#Deepest Root

Source: PAT A1021 Deepest Root (25 分) Description: 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…

PAT甲级题解分类byZlc

专题一字符串处理 A1001 Format(20) #include<cstdio> int main () { ]; int a,b,sum; scanf ("%d %d",&a,&b); sum=a+b; ) { printf ("-"); sum=-sum; } ; ) { s[top++]=; } ) { s[top++]=sum%; sum/=; } ;i>=;i--) { printf ("%d"…

PAT甲级1066. Root of AVL Tree

PAT甲级1066. Root of AVL Tree 题意: 构造AVL树,返回root点val. 思路: 了解AVL树的基本性质. AVL树 ac代码: C++ // pat1066.cpp : 定义控制台应用程序的入口点. // #include "stdafx.h" #include<iostream> #include<cstdio> #include<string> #include<algorithm> #include&l…

PAT 1021 Deepest Root[并查集、dfs][难]

1021 Deepest Root (25)(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…

[PAT] 1021 Deepest Root (25)（25 分）

1021 Deepest Root (25)(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…

pat甲级题解(更新到1013)

1001. A+B Format (20) 注意负数,没别的了. 用scanf来补前导0 和前导的空格很方便. #include <iostream> #include <cstdio> using namespace std; ]; int main() { int A,B; cin>>A>>B; A+=B; ) { A=-A; cout<<"-"; } ; while(A) { a[n++]=A%; A/=; } ;…

PAT甲级题解（慢慢刷中）

博主欢迎转载,但请给出本文链接,我尊重你,你尊重我,谢谢~http://www.cnblogs.com/chenxiwenruo/p/6102219.html特别不喜欢那些随便转载别人的原创文章又不给出链接的所以不准偷偷复制博主的博客噢~~ 时隔两年,又开始刷题啦,这篇用于PAT甲级题解,会随着不断刷题持续更新中,至于更新速度呢,嘿嘿,无法估计,不知道什么时候刷完这100多道题. 带*的是我认为比较不错的题目,其它的难点也顶多是细节处理的问题~ 做着做着,发现有些题目真的是太水了,都不想写题解了…

【转载】【PAT】PAT甲级题型分类整理

最短路径 Emergency (25)-PAT甲级真题(Dijkstra算法) Public Bike Management (30)-PAT甲级真题(Dijkstra + DFS) Travel Plan (30)-PAT甲级真题(Dijkstra + DFS,输出路径,边权) All Roads Lead to Rome (30)-PAT甲级真题-Dijkstra + DFS Online Map (30)-PAT甲级真题(Dijkstra + DFS) 最短路径扩展问题要求数最短路径有多…

PAT 甲级真题题解（1-62）

准备每天刷两题PAT真题.(一句话题解) 1001 A+B Format 模拟输出,注意格式 #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; string ans = ""; int main() { ; cin >> a >> b; c = a + b; ) {…

PAT甲级1127. ZigZagging on a Tree

PAT甲级1127. ZigZagging on a Tree 题意: 假设二叉树中的所有键都是不同的正整数.一个唯一的二叉树可以通过给定的一对后序和顺序遍历序列来确定.这是一个简单的标准程序,可以按顺序打印数字.但是,如果您认为问题太简单,那么您太天真了. 这次你应该以"锯齿形顺序"打印数字 - 也就是说,从根开始,逐级打印数字,从左到右交替,从右到左.例如,对于以下树,您必须输出:1 11 5 8 17 12 20 15. 输入规格: 每个输入文件包含一个测试用例.对于每种情况,第…

PAT甲级1123. Is It a Complete AVL Tree

PAT甲级1123. Is It a Complete AVL Tree 题意: 在AVL树中,任何节点的两个子树的高度最多有一个;如果在任何时候它们不同于一个,则重新平衡来恢复此属性.图1-4说明了旋转规则. 现在给出一系列插入, 您应该输出生成的AVL树的级别遍历序列,并告知它是否是完整的二叉树. 输入规格: 每个输入文件包含一个测试用例.对于每种情况,第一行包含正整数N(<= 20). 一行中的所有数字都以空格分隔. 输出规格: 对于每个测试用例,将键逐个插入到初始空的AVL树中.然后首先…

PAT甲级1119. Pre- and Post-order Traversals

PAT甲级1119. Pre- and Post-order Traversals 题意: 假设二叉树中的所有键都是不同的正整数.一个唯一的二进制树可以通过给定的一对后序和顺序遍历序列来确定,也可以通过预序和顺序遍历序列来确定.然而,如果仅给出了后序和预序遍历序列,则相应的树可能不再是唯一的. 现在给出一对postorder和preorder遍历序列,你应该输出树的相应的顺序遍历序列.如果树不是唯一的,只需输出任何一个树. 输入规格: 每个输入文件包含一个测试用例.对于每种情况,第一行给出正整数…

PAT甲级目录

树(23) 备注 1004 Counting Leaves 1020 Tree Traversals 1043 Is It a Binary Search Tree 判断BST,BST的性质 1053 Path of Equal Weight 1064 Complete Binary Search Tree 完全二叉树的顺序存储,BST的性质 1066 Root of AVL Tree 构建AVL树,模板题,需理解记忆 1079 Total Sales of Supply Chain…

PAT 甲级真题

1019. General Palindromic Number 题意:求数N在b进制下其序列是否为回文串,并输出其在b进制下的表示. 思路:模拟N在2进制下的表示求法,“除b倒取余”,之后判断是否回文. #include<iostream> #include<cstdio> using namespace std; ]; int main() { int n, b; scanf("%d%d", &n, &b); , tmp = n,yu=tmp…

1021. Deepest Root (25)——DFS+并查集

http://pat.zju.edu.cn/contests/pat-a-practise/1021 无环连通图也可以视为一棵树,选定图中任意一点作为根,如果这时候整个树的深度最大,则称其为 deepest root. 给定一个图,按升序输出所有 deepest root.如果给定的图有多个连通分量,则输出连通分量的数量. 1.使用并查集判断图是否为连通的. 2.任意选取一点,做 dfs 搜索,选取其中一个最远距离的点 A,再做一次 dfs,找到的所有距离最远的点以及点 A 都是 deepest…