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. word的"bug"

    发表博客发现,从word复制文本到chrome浏览器上的博客时, 如果复制完后立即关闭word,那么将无法粘贴到通过chrome浏览器访问的博客上,也无法粘贴到记事本上: 但是复制完立即关闭word后 ...

  2. socket基础编程-1

    server端和client端 1.server端: import socket server=socket.socket() server.bind(('localhost',8080)) serv ...

  3. element-ui 源码解析 二

    Carousel 走马灯源码解析 1. 基本原理:页面切换 页面切换使用的是 transform 2D 转换和 transition 过渡 可以看出是采用内联样式来实现的 举个栗子 <div : ...

  4. 关于mysql远程登录问题

    问题:mysql不能实现远程登录 前提:mysql开启了远程登录账号,安全组也放行了3306,防火墙是iptables,也加入了3306放行,但是还是不能实现远程访问 解决办法,使用iptables ...

  5. java sort排序原理

    事实上Collections.sort方法底层就是调用的Arrays.sort方法,而Arrays.sort使用了两种排序方法,快速排序和优化的归并排序. 快速排序主要是对那些基本类型数据(int,s ...

  6. 莫烦theano学习自修第三天【共享变量】

    1. 代码实现 #!/usr/bin/env python #! _*_ coding:UTF-8 _*_ import numpy as np import theano.tensor as T i ...

  7. Linux中,去掉终端显示的当前目录的绝对路径

    Linux中,去掉终端显示的当前目录的绝对路径 去~/.bashrc中,找到PS1变量的定义,如果没有,手动加上: 可以将显示输出到标题栏上: #export PS1="[e]2;u@H w ...

  8. Maven自动部署jar包到Neuxs

      1. 修改maven配置(setting.xml) 添加neuxs的用户名和密码: <server> <id>my-deploy-release</id> &l ...

  9. Lodop打印设计(PRINT_DESIGN)介绍

    打印设计(PRINT_DESIGN)界面上方有两栏菜单栏,举例说明(文本框,条码,图形等).(1)第一排最左侧第一个功能,位置移动:控制里面元素微上下左右移动,每次移动一个px.(用于微调,普通调整可 ...

  10. 定位linux jdk安装路径

    如何在一台Linux服务器上查找JDK的安装路径呢? 有那些方法可以查找定位JDK的安装路径?是否有一些局限性呢? 下面总结了一下如何查找JDK安装路径的方法. 1:echo $JAVA_HOME 使 ...