A1147. Heaps
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
- 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 ...
- PAT_A1147#Heaps
Source: PAT A1147 Heaps (30 分) Description: In computer science, a heap is a specialized tree-based ...
- 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 ...
- 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& ...
- 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 ...
随机推荐
- 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> ...
- 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: ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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( ...