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
- js如何实现上拉加载更多...
我们在项目中经常使用到下拉加载更多,之前要么是底部写加载按钮,要么是引入插件.今天终于有时间手写一个了,之前感觉挺麻烦,明白原理后,其实很简单... scrollTop:滚动视窗的高度距离window ...
- Codeforces - 1198C - Matching vs Independent Set - 贪心
https://codeforces.com/contest/1198/problem/C 要选取一个大小大于等于n的匹配或者选取一个大小大于等于n的独立集. 考虑不断加入匹配集,最终加入了x条边. ...
- P5504 [JSOI2011]柠檬
传送门 显然考虑 $dp$ ,发现从右往左和从左往右是一样的,所以只考虑一边就行 发现对于切的左右端点,选择的 $s0$ 一定要为左右端点的贝壳大小,不然这个端点不产生贡献还不如分开来单个贡献 所以设 ...
- 09.Linux系统由于不正常关机导致的分区问题
问题:Error:UNEXPECTED INCONSISTENCY: RUN fsck MANUALLY Give root password for maintenance ------------ ...
- linux --memcached的安装与配置
转载:http://blog.sina.com.cn/s/blog_4829b9400101piil.html 1.准备安装包:libevent-2.0.21-stable.tar.gz 和memca ...
- SOAP、WSDL、 UDDI之间的关系
SOAP(Simple Object Access Protocol) 简单对象访问协议: WSDL(Web Services Description Language) Web服务描述语言: UDD ...
- Java中Comparable接口和Comparator接口的简单用法
对象比较器 1.Comparable接口 此接口强行对实现它的每个类的对象进行整体排序,这种排序成为类的自然排序,类的compareTo方法称为类的自然比较方法. 代码示例 import java.u ...
- Test 6.24 T2 集合
问题描述 有一个可重集合,一开始只有一个元素 0. 你可以进行若干轮操作,每轮你需要对集合中每个元素 x 执行以下三种操作之一: 将 x 变为 x+1; 选择两个非负整数 y,z 满足 y+z=x , ...
- Netty模型
- JS自定义随机键盘
<!DOCTYPE html> <html lang="en"> <head> <meta charset="UTF-8&quo ...