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
#include<iostream>
#include<cstdio>
using namespace std;
int N, M, tree[];
int isMaxheap(){
for(int k = N / ; k >= ; k--){
int max;
if(*k + <= N && tree[*k+] > tree[*k]){
max = tree[*k + ];
}else{
max = tree[*k];
}
if(tree[k] < max)
return ;
}
return ;
}
int isMinheap(){
for(int k = N / ; k >= ; k--){
int min;
if(*k+ <= N && tree[*k+] < tree[*k]){
min = tree[*k+];
}else min = tree[*k];
if(tree[k] > min)
return ;
}
return ;
}
int cnt = ;
void postOrder(int root){
if(root > N)
return;
postOrder(root*);
postOrder(root* + );
cnt++;
if(cnt == N)
printf("%d\n", tree[root]);
else printf("%d ", tree[root]);
}
int main(){
scanf("%d%d", &M, &N);
for(int i = ; i < M; i++){
for(int i = ; i <= N; i++){
scanf("%d", &tree[i]);
}
if(isMaxheap()){
printf("Max Heap\n");
}else if(isMinheap()){
printf("Min Heap\n");
}else{
printf("Not Heap\n");
}
cnt = ;
postOrder();
}
cin >> N;
return ;
}

总结:

1、题意:检查每个给出的序列是否是一个大根堆或者小根堆。

A1147. Heaps的更多相关文章

  1. PAT A1147 Heaps (30 分)——完全二叉树,层序遍历,后序遍历

    In computer science, a heap is a specialized tree-based data structure that satisfies the heap prope ...

  2. PAT_A1147#Heaps

    Source: PAT A1147 Heaps (30 分) Description: In computer science, a heap is a specialized tree-based ...

  3. PAT (Advanced Level) Practice(更新中)

    Source: PAT (Advanced Level) Practice Reference: [1]胡凡,曾磊.算法笔记[M].机械工业出版社.2016.7 Outline: 基础数据结构: 线性 ...

  4. 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& ...

  5. CodeForces 353B Two Heaps

    B. Two Heaps   Valera has 2·n cubes, each cube contains an integer from 10 to 99. He arbitrarily cho ...

  6. 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 ...

  7. 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 ...

  8. 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 ...

  9. 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 ...

随机推荐

  1. MySQL执行语句的顺序

    MySQL的语句一共分为11步,最先执行的总是FROM操作,最后执行的是LIMIT操作.其中每一个操作都会产生一张虚拟的表,这个虚拟的表作为一个处理的输入,只是这些虚拟的表对用户来说是透明的,但是只有 ...

  2. SpringBoot之处理JSON数据举例

    SpringBoot使用@RequestBody注解会自动将请求body中的json数据绑定到参数上.使用@ResponseBody注解,在返回参数时自动将对象转换为JSON格式返回. 举例代码: c ...

  3. 获取网络图片并显示在picturbox上,byte[]数组转换成Image:

    private void getWebPicture_Click(object sender, EventArgs e) { WebRequest request = WebRequest.Creat ...

  4. linux寻找依赖文件

    在linux下编译安装软件有时候会遇到依赖文件找不到的情况,很多时候可以通过 sudo apt install -f 来解决:实在找不到怎么办,还有一个绝招可以用: 安装  apt-file sudo ...

  5. 今天开始学Pattern Recognition and Machine Learning (PRML),章节5.2-5.3,Neural Networks神经网络训练(BP算法)

    转载请注明出处:http://www.cnblogs.com/xbinworld/p/4265530.html 这一篇是整个第五章的精华了,会重点介绍一下Neural Networks的训练方法——反 ...

  6. asp.net core mvc ajaxform submit files

    <form id="form1" method="post" enctype="multipart/form-data" asp-co ...

  7. java web 开发入门 --- tomcat/servlet/jsp

    在做java web 开发时,要先安装tomcat.它是一个web服务器,也叫web容器,我们把写好的jsp, html页面放到它里面,然后启动它,就可以用浏览器访问这些页面,地址栏中输入localh ...

  8. hdu 5652(并查集)

    题意:很久之前,在中国和印度之间有通路,通路可以简化为一个n*m的字符串,0表示能通过,1表示障碍,每过一年就有一个坐标变成1,问你什么时候,通路彻底无法通过: 解题思路:无向图的连通性,一般直接搜索 ...

  9. url.openconnection() 设置超时时间

    System.setProperty("sun.net.client.defaultConnectTimeout", "30000"); System.setP ...

  10. 【题解】K乘积

    题目描述 有N个数,每个数的范围是[-50,50],现在你要从这N个数中选出K个,使得这K个数的乘积最大. 输入格式 第一行,N和K. 1 <= N <= 50.  1 <= K & ...