PTA 1155 Heap Paths (DFS)
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))
One thing for sure is that all the keys along any path from the root to a leaf in a max/min heap must be in non-increasing/non-decreasing order.
Your job is to check every path in a given complete binary tree, in order to tell if it is a heap or not.
Input Specification:
Each input file contains one test case. For each case, the first line gives a positive integer \(N (1<N≤1,000)\), the number of keys in the tree. Then the next line 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, first print all the paths from the root to the leaves. Each path occupies a line, with all the numbers separated by a space, and no extra space at the beginning or the end of the line. The paths must be printed in the following order: for each node in the tree, all the paths in its right subtree must be printed before those in its left subtree.
Finally print in a line
Max Heapif it is a max heap, orMin Heapfor a min heap, orNot Heapif it is not a heap at all.
Sample Input 1:
8
98 72 86 60 65 12 23 50
Sample Output 1:
98 86 23
98 86 12
98 72 65
98 72 60 50
Max Heap
Sample Input 2:
8
8 38 25 58 52 82 70 60
Sample Output 2:
8 25 70
8 25 82
8 38 52
8 38 58 60
Min Heap
Sample Input 3:
8
10 28 15 12 34 9 8 56
Sample Output 3:
10 15 8
10 15 9
10 28 34
10 28 12 56
Not Heap
题意
给定一个长度为 \(N\) 的数组,输出从根节点到叶子结点的每一条路径,并且判断是否是堆。
思路
直接 dfs 输出路径。设立两个变量 Max 和 Min 统计父节点比子节点大的个数和父节点比子节点小的个数。如果两者都不为 0,说明不是堆;如果 Max 为 0,说明是小顶堆,如果 Min 为 0,说明是大顶堆。
代码
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e3 + 10;
int arr[maxn];
int n;
int path[11]; // 保存路径
int Max = 0, Min = 0;
void dfs(int step, int id) {
path[step] = arr[id];
if(id * 2 > n) {
for(int i = 1; i < step; ++i) {
cout << path[i] << " ";
}
cout << path[step] << endl;
return;
}
int l = id * 2, r = id * 2 + 1; // 左右儿子结点
if(r <= n) {
if(path[step] < arr[r]) ++Min;
if(path[step] > arr[r]) ++Max;
dfs(step + 1, r);
}
if(l <= n) {
if(path[step] < arr[l]) ++Min;
if(path[step] > arr[l]) ++Max;
dfs(step + 1, l);
}
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
cin >> n;
for(int i = 1; i <= n; ++i) {
cin >> arr[i];
}
dfs(1, 1);
if(Min == 0) {
cout << "Max Heap" << endl;
} else if(Max == 0) {
cout << "Min Heap" << endl;
} else {
cout << "Not Heap" << endl;
}
return 0;
}
PTA 1155 Heap Paths (DFS)的更多相关文章
- PAT Advanced 1155 Heap Paths (30 分)
In computer science, a heap is a specialized tree-based data structure that satisfies the heap prope ...
- PAT Advanced 1155 Heap Paths (30) [DFS, 深搜回溯,堆]
题目 In computer science, a heap is a specialized tree-based data structure that satisfies the heap pr ...
- 1155 Heap Paths (30 分)(堆+dfs遍历)
比较简单的一题 遍历左右的时候注意一下 #include<bits/stdc++.h> using namespace std; ; ]; ; vector<int>t; ve ...
- PAT 甲级 1155 Heap Paths
https://pintia.cn/problem-sets/994805342720868352/problems/1071785408849047552 In computer science, ...
- pat甲级 1155 Heap Paths (30 分)
In computer science, a heap is a specialized tree-based data structure that satisfies the heap prope ...
- 1155 Heap Paths
题干前半略. Sample Input 1: 8 98 72 86 60 65 12 23 50 Sample Output 1: 98 86 23 98 86 12 98 72 65 98 72 ...
- PAT甲级 1155 Heap Paths (30分) 堆模拟
题意分析: 给出一个1000以内的整数N,以及N个整数,并且这N个数是按照完全二叉树的层序遍历输出的序列,输出所有的整条的先序遍历的序列(根 右 左),以及判断整棵树是否是符合堆排序的规则(判断是大顶 ...
- PAT_A1155#Heap Paths
Source: PAT A1155 Heap Paths (30 分) Description: In computer science, a heap is a specialized tree-b ...
- PAT A1155 Heap Paths (30 分)——完全二叉树,层序遍历,特定dfs遍历
In computer science, a heap is a specialized tree-based data structure that satisfies the heap prope ...
随机推荐
- Super Mario HDU 4417 主席树区间查询
Super Mario HDU 4417 主席树区间查询 题意 给你n个数(编号从0开始),然后查询区间内小于k的数的个数. 解题思路 这个可以使用主席树来处理,因为这个很类似查询区间内的第k小的问题 ...
- RMAN备份与恢复 —— 完全恢复与不完全恢复
名词解释: 顾名思义,完全恢复就是指数据没有丢失的恢复了.不完全恢复是指恢复后有部分数据丢失.它们是数据库的两种恢复方式. 完全恢复:利用重做日志或增量备份将数据块恢复到最接近当前时间的 ...
- Aurora测试----随机数字产生
在xilinx模板中,存在一个Aurora样本工程,包含众多的子函数,本系列本文将逐一对其进行解析,首先是aurora_8b10b_0_FRAME_GEN函数,根据官方的说明,其作用是:该模块是一个模 ...
- C# Xml.Serialization 节点重命名
XmlElement 节点重命名 XmlRoot 根节点重名称 XmlArray List集合添加根节点 XmlArrayItem List集合中子节点重命名 [Serializable] 将该类标记 ...
- SQL SERVER 数据库跨服务器备份
原文:https://www.cnblogs.com/jaday/p/6088200.html 需求介绍:每天备份线上正式库并且把备份文件复制到测试服务器,测试服务器自动把数据库备份文件还原. 方案介 ...
- ajax图片上传(asp.net +jquery+ashx)
一.建立Default.aspx页面 <%@ Page Language="C#" AutoEventWireup="true" CodeFile=&q ...
- python面向对象--反射机制
class Black: feture="ugly" def __init__(self,name,addr): self.addr=addr self.name=name def ...
- mysql的mod函数
取余是用函数mod(numer1,number2),其返回的值为其余数值 如:mod(id,2) = 1 返回id号是奇数的id
- 牛客CSP-S提高模拟4 赛后总结
前言 其实前面已经打了 3 场牛客 3 场计蒜客的比赛,都没有写总结,今天先提一下以前的情况 计蒜客 1 :0+0+0 = 0 (心态崩了,写挂了) 牛客 1: 0+0+0 = 0 (T1博弈论,T2 ...
- bzoj3809 Gty的二逼妹子序列 & bzoj3236 [Ahoi2013]作业 莫队+分块
题目传送门 https://lydsy.com/JudgeOnline/problem.php?id=3809 https://lydsy.com/JudgeOnline/problem.php?id ...