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. vue页面是否缓存的两种方式

    第一种 <keep-alive> <router-view v-if="$route.meta.keepAlive"></router-view> ...

  2. python 获取列表中次大的数值.

    需求: 1.写个函数,把一组数字传到函数中,然后取出最大值和次大值. 2.不能使用排序函数. 分析: Q: list = [100,50,60,70,30,45] 怎么从这个列表中取出最大值? A: ...

  3. Day 4-9 subprocess模块

    我们经常需要通过Python去执行一条系统命令或脚本,系统的shell命令是独立于你的python进程之外的,每执行一条命令,就是发起一个新进程,通过python调用系统命令或脚本的模块在python ...

  4. Navicat 远程连接Docker容器中的mysql 报错:1251 - Client does not support authentication protocol 解决办法。

    出现这个问题 首先进入 1.docker exec -it mysql02 bash      //mysql02是mysql容器的别名 2.mysql -uroot -p 3.输入密码 4.进入my ...

  5. Mybatis核心配置文件SqlMapConfig.xml

    配置内容: SqlMapConfig.xml中配置的内容和顺序如下: 1.properties(属性) 2.settings(全局配置参数) 3.typeAliases(类型别名) 4.typeHan ...

  6. 异步httpclient(httpasyncclient)的使用与总结

    参考:异步httpclient(httpasyncclient)的使用与总结 1. 前言应用层的网络模型有同步与异步.同步意味当前线程是阻塞的,只有本次请求完成后才能进行下一次请求;异步意味着所有的请 ...

  7. Maven 项目 无缘无故报错:版本冲突,其他机器上正常-提交的时候报冲突怎么也解决不掉

    2018年: maven突然之间报错了,显示版本冲突,但是其他的机器是好的, 使用命令:mvn compile -P dev -e; 看看测试环境有没有问题,还是有问题.而且,刚开始只是报错:erro ...

  8. python数据结构与算法第十四天【二分查找】

    1.二分查找的原理 对于已经排序的列表进行最快速度的查找 2. 代码实现 (1)递归实现 def binary_search(alist, item): if len(alist) == 0: ret ...

  9. python数据结构与算法第十三天【归并排序】

    1.代码实现 def merge_sort(alist): if len(alist) <= 1: return alist # 二分分解 num = len(alist)/2 left = m ...

  10. Ajax的post表单,不在url后接一大串参数键值对的方法

    $('#loginForm').on('submit',function (ev) { //阻止表单参数附在url后面 ev.stopPropagation(); ev.preventDefault( ...