PAT_A1147#Heaps
Source:
Description:
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, orMin Heap
for a min heap, orNot 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
Keys:
Attention:
- 有点水了哈0,0;
Code:
/*
Data: 2019-06-29 16:15:43
Problem: PAT_A1147#Heaps
AC: 19:20 题目大意:
判断给定的完全二叉树是否是堆
输入:
第一行给出测试数M(<=100)和结点数N[1,1e3]
接下来N行,逐层给出完全二叉树的各个键值
输出:
大根堆,小根堆,非堆;
接下来输出二叉树的后序遍历 基本思路:
静态树存储BST,遍历判断是否为堆
*/
#include<cstdio>
#include<vector>
using namespace std;
const int M=1e3+;
int bst[M],Min,Max,n,m;
vector<int> path; void Travel(int root)
{
if(root > n)
return;
if(root!=)
{
if(bst[root/] > bst[root])
Min=;
if(bst[root/] < bst[root])
Max=;
}
Travel(root*);
Travel(root*+);
path.push_back(bst[root]);
} int main()
{
#ifdef ONLINE_JUDGE
#else
freopen("Test.txt", "r", stdin);
#endif // ONLINE_JUDGE scanf("%d%d", &m,&n);
while(m--)
{
for(int i=; i<=n; i++)
scanf("%d", &bst[i]);
Min=;
Max=;
path.clear();
Travel();
if(Max) printf("Max Heap\n");
else if(Min) printf("Min Heap\n");
else printf("Not Heap\n");
for(int i=; i<n; i++)
printf("%d%c", path[i],i==n-?'\n':' ');
} return ;
}
PAT_A1147#Heaps的更多相关文章
- CodeForces 353B Two Heaps
B. Two Heaps Valera has 2·n cubes, each cube contains an integer from 10 to 99. He arbitrarily cho ...
- DSP\BIOS调试Heaps are enabled,but not set correctly
转自:http://blog.sina.com.cn/s/blog_735f291001015t9i.html Heaps are enabled, but the segment for DSP/B ...
- CSU 1616: Heaps(区间DP)
题目链接:http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1616 1616: Heaps Time Limit: 2 Sec Memory Lim ...
- Codeforces Round #300 F - A Heap of Heaps (树状数组 OR 差分)
F. A Heap of Heaps time limit per test 3 seconds memory limit per test 512 megabytes input standard ...
- Heaps(Contest2080 - 湖南多校对抗赛(2015.05.10)(国防科大学校赛决赛-Semilive)+scu1616)
Problem H: Heaps Time Limit: 2 Sec Memory Limit: 128 MBSubmit: 48 Solved: 9[Submit][Status][Web Bo ...
- Fibonacci Heaps
Mergeable heapsA mergeable heap is any data structure that supports the following five operations,in ...
- PAT 1147 Heaps[难]
1147 Heaps(30 分) In computer science, a heap is a specialized tree-based data structure that satisfi ...
- [PAT] 1147 Heaps(30 分)
1147 Heaps(30 分) In computer science, a heap is a specialized tree-based data structure that satisfi ...
- Codeforces 538 F. A Heap of Heaps
\(>Codeforces \space 538 F. A Heap of Heaps<\) 题目大意 :给出 \(n\) 个点,编号为 \(1 - n\) ,每个点有点权,将这些点构建成 ...
随机推荐
- 开源GIS软件 4
空间数据操作框架 Apache SIS Apache SIS 是一个空间的框架,可以更好地搜索,数据聚类,归档,或任何其他相关的空间坐标表示的需要. kvwmap kvwmap是一个采用PHP开发的W ...
- Eclipse ADT 导入别的电脑开发的项目
用Eclipse开发的时候常常要导入别的电脑开发的项目,常常会出错,甚至导入不了. 方法一: 把你正在使用的Eclipse开发的随便一个项目.打开,把下图这三个文件复制过去你要导入的项目.覆盖.然后再 ...
- HDU 2563 统计问题(递推)
题目链接:http://acm.hdu.edu.cn/showproblem.php? pid=2563 将向上移的步数设为a[n],将向左右移的步数设为b[n],有a[n]=a[n-1]+b[n-1 ...
- shell EOF注意点
当sqlplus与shell交互的时候我们这么用 su - oracle -c "sqlplus / as sysdba<<EOF select * from gv($insta ...
- jenkins+jmeter+ant+jmeter在Jenkins上报告
1.jmeter+ant 参考 http://www.cnblogs.com/dieyaxianju/p/8268802.html 2.在jenkins上配置 3.执行成功 4.配置报告 参考 下载 ...
- 使用百度地图API进行Android地图应用开发(Eclipse)
随着基于位置的服务的兴起,地图类App呈现爆发趋势.随着而来的是地图供应商开放大量的API.供开发人员开发基于PC或者移动端的应用程序. 如今我们研究使用百度地图SDK进行Android项目的开发. ...
- Mybatis 框架文档 超具体笔记
1 Mybatis入门 1.1 单独使用jdbc编程问题总结 1.1.1 jdbc程序 Public static void main(String[] args) { Connec ...
- Android系统Recovery工作原理之使用update.zip升级过程分析(七)---Recovery服务的核心install_package函数【转】
本文转载自:http://blog.csdn.net/mu0206mu/article/details/7465514 一. Recovery服务的核心install_package(升级 ...
- 2749: [HAOI2012]外星人
首先像我一样把柿子画出来或者看下hint 你就会发现其实是多了个p-1这样的东东 然后除非是2他们都是偶数,而2就直接到0了 算一下2出现的次数就好 #include<cstdio> #i ...
- 生活的 tricks
1. 远距离传递 传真(需要附近有传真机):发 QQ.微信拍照,自己打印: 2. 超市的设计 如果是两层的话,入口一定在第一楼,出口在第二楼,也即当你需要出去的时候,需要贯穿整个超市: 用的在第一楼: ...