[PAT] 1147 Heaps(30 分)
1147 Heaps(30 分)
In computer science, a heap is a specialized tree-based data structure that satisfies the heap property: if P is a parent node of C, then the key (the value) of P is either greater than or equal to (in a max heap) or less than or equal to (in a min heap) the key of C. A common implementation of a heap is the binary heap, in which the tree is a complete binary tree. (Quoted from Wikipedia at https://en.wikipedia.org/wiki/Heap_(data_structure)
Your job is to tell if a given complete binary tree is a heap.
Input Specification:
Each input file contains one test case. For each case, the first line gives two positive integers: M (≤ 100), the number of trees to be tested; and N (1 < N ≤ 1,000), the number of keys in each tree, respectively. Then M lines follow, each contains N distinct integer keys (all in the range of int), which gives the level order traversal sequence of a complete binary tree.
Output Specification:
For each given tree, print in a line Max Heap if it is a max heap, or Min Heap for a min heap, or Not Heap if it is not a heap at all. Then in the next line print the tree's postorder traversal sequence. All the numbers are separated by a space, and there must no extra space at the beginning or the end of the line.
Sample Input:
3 8
98 72 86 60 65 12 23 50
8 38 25 58 52 82 70 60
10 28 15 12 34 9 8 56
Sample Output:
Max Heap
50 60 65 72 12 23 86 98
Min Heap
60 58 52 38 82 70 25 8
Not Heap
56 12 34 28 9 8 15 10
题意:
给定层序遍历的树,判断是否为大小堆。
思路:
先将输出转化为一棵树,这样就好做了,就是比较暴力
看看 柳婼 の blog 是怎么做的
题解:
#include<cstdlib>
#include<cstdio>
#include<vector>
using namespace std;
struct node {
int value;
int left;
int right;
};
bool isMaxHeap(vector<node> nodes, int now) {
bool isMax = true;
< nodes.size()) {
].value) {
isMax = false;
}
else {
)) {
isMax = false;
}
}
}
+ < nodes.size()) {
+ ].value) {
isMax = false;
}
else {
+ )) {
isMax = false;
}
}
}
return isMax;
}
bool isMinHeap(vector<node> nodes, int now) {
bool isMin = true;
< nodes.size()) {
].value) {
isMin = false;
}
else {
)) {
isMin = false;
}
}
}
+ < nodes.size()) {
+ ].value) {
isMin = false;
}
else {
+ )) {
isMin = false;
}
}
}
return isMin;
}
void PostOrder(vector<node> nodes, int now) {
< nodes.size()) {
PostOrder(nodes, now * );
}
+ < nodes.size()) {
PostOrder(nodes, now * + );
}
//nodes[nodes.size() - 1]这个结点不一定是先输出的。
printf( ? "\n" : " ");
}
int main() {
int n, m;
scanf("%d %d", &n, &m);
; i < n; i++) {
vector<node> nodes(m+);
; j <= m; j++) {
scanf("%d", &nodes[j].value);
nodes[j].left = j * ;
nodes[j].right = j * + ;
}
)) {
printf("Max Heap\n");
}
)) {
printf("Min Heap\n");
}
else {
printf("Not Heap\n");
}
PostOrder(nodes, );
}
;
}
[PAT] 1147 Heaps(30 分)的更多相关文章
- PAT 甲级 1147 Heaps (30 分) (层序遍历,如何建树,后序输出,还有更简单的方法~)
1147 Heaps (30 分) In computer science, a heap is a specialized tree-based data structure that sati ...
- PAT 1147 Heaps[难]
1147 Heaps(30 分) In computer science, a heap is a specialized tree-based data structure that satisfi ...
- PAT Advanced 1147 Heaps (30) [堆,树的遍历]
题目 In computer science, a heap is a specialized tree-based data structure that satisfies the heap pr ...
- PAT 1147 Heaps
https://pintia.cn/problem-sets/994805342720868352/problems/994805342821531648 In computer science, a ...
- 1147. Heaps (30)
In computer science, a heap is a specialized tree-based data structure that satisfies the heap prope ...
- PAT甲级:1064 Complete Binary Search Tree (30分)
PAT甲级:1064 Complete Binary Search Tree (30分) 题干 A Binary Search Tree (BST) is recursively defined as ...
- PAT A1127 ZigZagging on a Tree (30 分)——二叉树,建树,层序遍历
Suppose that all the keys in a binary tree are distinct positive integers. A unique binary tree can ...
- 【PAT】1053 Path of Equal Weight(30 分)
1053 Path of Equal Weight(30 分) Given a non-empty tree with root R, and with weight Wi assigned t ...
- 【PAT】1091 Acute Stroke(30 分)
1091 Acute Stroke(30 分) One important factor to identify acute stroke (急性脑卒中) is the volume of the s ...
随机推荐
- Hadoop,大数据,云计算三者之间的关系
大数据和云计算是何关系?关于大数据和云计算的关系人们通常会有误解.而且也会把它们混起来说,分别做一句话直白解释就是:云计算就是硬件资源的虚拟化;大数据就是海量数据的高效处理.大数据.hadoop及云计 ...
- Indexing GROUP BY
SQL databases use two entirely different group by algorithms. The first one, the hash algorithm, agg ...
- STL使用总结
转载于http://blog.csdn.net/daisy_chenting/article/details/6898184 1. 概述 泛型编程思想最早缘于A.Stepanov提出的部分算法可 ...
- 8VC Venture Cup 2016 - Final Round (Div. 2 Edition) A
A. Orchestra time limit per test 2 seconds memory limit per test 256 megabytes input standard input ...
- mysql绿色版安装,多实例安装
1.为什么要装多个mysql多实例? 关于这个的原因,我目前了解为建立一个主数据库,一个或者多个从库,实现一主多从或者主从复制的目的. 2.设么是mysql的多实例? MySQL多实例就是在一台机器上 ...
- Android提示框与通知的使用
1.通知 Android 3.0以前使用的方法 NotificationManager nm = (NotificationManager) getSystemService(NOTIFICATION ...
- \(\rm LightOJ 1371 - Energetic Pandas 简单计数+组合\)
http://www.lightoj.com/volume_showproblem.php?problem=1371 题意:给你n根竹子,和n只熊猫(XD),每个熊猫只能选择重量不大于它的竹子,问有几 ...
- 20155335俞昆《java程序设计》第三周总结
20155335 2006-2007-2 <Java程序设计>第三周学习总结 ## 教材学习内容总结 首先,关键是区基本类型和类类型,,产生对象必须定义类,类是一个概念,并不存在,对 ...
- Java并发——关键字synchronized解析
synchronized用法 在Java中,最简单粗暴的同步手段就是synchronized关键字,其同步的三种用法: ①.同步实例方法,锁是当前实例对象 ②.同步类方法,锁是当前类对象 ③.同步代码 ...
- 【洛谷 T47488】 D:希望 (点分治)
题目链接 看到这种找树链的题目肯定是想到点分治的. 我码了一下午,\(debug\)一晚上,终于做到只有两个点TLE了. 我的是不完美做法 加上特判\(A\)了这题qwq 记录每个字母在母串中出现的所 ...