PAT甲级——1147 Heaps【30】
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
第一种方法,比较笨,重建整棵树,然后判断是否时大根堆和小根堆,然后再遍历出后序遍历
#include <iostream>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std;
int n, m;
vector<int>level, post;
struct Node
{
int val;
Node *l, *r;
Node(int a = ) :val(a), l(nullptr), r(nullptr) {}
};
Node* creatTree(bool &flag, const bool isMax)
{
Node* root = new Node(level[]);
int k = ;
queue<Node*>q;
q.push(root);
while (k < m)
{
Node *p = q.front();
q.pop();
p->l = new Node(level[k++]);
if (isMax && p->val<p->l->val || !isMax && p->val>p->l->val)
flag = false;
q.push(p->l);
if (k >= m)break;
p->r = new Node(level[k++]);
if (isMax && p->val < p->r->val || !isMax && p->val > p->r->val)
flag = false;
q.push(p->r);
}
return root;
}
void postOrder(Node *root)
{
if (root == nullptr)
return;
postOrder(root->l);
postOrder(root->r);
post.push_back(root->val);
}
int main()
{
cin >> n >> m;
while (n--)
{
level.clear();
level.resize(m);
post.clear();
int minN = INT32_MAX, maxN = -;
for (int i = ; i < m; ++i)
{
cin >> level[i];
minN = minN < level[i] ? minN : level[i];
maxN = maxN > level[i] ? maxN : level[i];
}
bool flag = true, isMax = false;
Node *root = nullptr;
if (level[] == minN)//小根堆
{
isMax = false;
root = creatTree(flag, isMax);
}
else if (level[] == maxN)
{
isMax = true;
root = creatTree(flag, isMax);
}
else
{
flag = false;
root = creatTree(flag, isMax);
}
postOrder(root);
if (flag && isMax)
printf("Max Heap\n");
else if (flag && !isMax)
printf("Min Heap\n");
else
printf("Not Heap\n");
for (int i = ; i < m; ++i)
cout << (i == ? "" : " ") << post[i];
cout << endl;
}
return ;
}
第二种方法,简单点,通过完全二叉树的性质,直接判断并得出后序遍历结果
#include <iostream>
#include <vector>
using namespace std;
int n, m;
vector<int>level, post;
void postOrder(int index)
{
if (index >= m)return;
postOrder(index * + );
postOrder(index * + );
post.push_back(level[index]);
}
int main()
{
cin >> n >> m;
while (n--)
{
level.resize(m);
for (int i = ; i < m; ++i)
cin >> level[i];
bool isMaxHeap = level[] >= level[] ? true : false;
bool flag = true;
for (int i = ; i < (m - ) / && flag; ++i)
{
int L = i * + , R = i * + ;
if (isMaxHeap && (level[i] < level[L] || R < m && level[i] < level[R]))
flag = false;
if (!isMaxHeap && (level[i] > level[L] || R<m && level[i] > level[R]))
flag = false;
}
if (flag && isMaxHeap)
printf("Max Heap\n");
else if (flag && !isMaxHeap)
printf("Min Heap\n");
else
printf("Not Heap\n");
postOrder();
for (int i = ; i < m; ++i)
cout << (i == ? "" : " ") << post[i];
cout << endl;
}
return ;
}
PAT甲级——1147 Heaps【30】的更多相关文章
- PAT 甲级 1147 Heaps (30 分) (层序遍历,如何建树,后序输出,还有更简单的方法~)
1147 Heaps (30 分) In computer science, a heap is a specialized tree-based data structure that sati ...
- PAT Advanced 1147 Heaps (30) [堆,树的遍历]
题目 In computer science, a heap is a specialized tree-based data structure that satisfies the heap pr ...
- PAT 甲级1057 Stack (30 分)(不会,树状数组+二分)*****
1057 Stack (30 分) Stack is one of the most fundamental data structures, which is based on the prin ...
- pat 甲级 1057 Stack(30) (树状数组+二分)
1057 Stack (30 分) Stack is one of the most fundamental data structures, which is based on the princi ...
- 1147. Heaps (30)
In computer science, a heap is a specialized tree-based data structure that satisfies the heap prope ...
- PAT甲级 堆 相关题_C++题解
堆 目录 <算法笔记>重点摘要 1147 Heaps (30) 1155 Heap Paths (30) <算法笔记> 9.7 堆 重点摘要 1. 定义 堆是完全二叉树,树中每 ...
- [PAT] 1147 Heaps(30 分)
1147 Heaps(30 分) In computer science, a heap is a specialized tree-based data structure that satisfi ...
- PAT 1147 Heaps[难]
1147 Heaps(30 分) In computer science, a heap is a specialized tree-based data structure that satisfi ...
- PAT甲级:1064 Complete Binary Search Tree (30分)
PAT甲级:1064 Complete Binary Search Tree (30分) 题干 A Binary Search Tree (BST) is recursively defined as ...
随机推荐
- MySQL 同一字段匹配多个值
delete from api_log WHERE curl LIKE '%.css' or curl LIKE '%.js' or curl LIKE '%.JPG'; 删除字段curl以 js,c ...
- Shell脚本中判断输入变量或者参数是否为空的方法
shell判断一个变量是否为空方法总结 https://www.jb51.net/article/154835.htm 1.判断变量 复制代码代码如下: read -p "input a w ...
- addr2line探秘 [從ip讀出程式中哪行出錯]
addr2line探秘 在Linux下写C/C++程序的程序员,时常与Core Dump相见.在内存越界访问,收到不能处理的信号,除零等错误出现时,我们精心或不精心写就的程序就直接一命呜呼了,Core ...
- rpm相关
rpm命令是RPM软件包的管理工具.rpm原本是Red Hat Linux发行版专门用来管理Linux各项套件的程序,由于它遵循GPL规则且功能强大方便,因而广受欢迎.逐渐受到其他发行版的采用.RPM ...
- JS事件委托(事件代理,dom2级事件)
一.前言 说实话,真问我什么是事件委托,我肯定gg,还好查了一下,原来就是我之前练习过的DOM2级事件的应用. 二.什么是事件委托? 事件委托就是当事件触发时,把要做的事委托给父元素(或父元素的父元素 ...
- jQuery封装轮播图插件
// 布局要求,必须有一个容器,图片和两个按钮,布局方式自定,小圆点样式固定 // <div class="all"> // <img src="img ...
- CocoaPods CDN: trunk Repo update failed
问题 今天升级 CocoaPods 到 1.8.4 版本但是随即问题就来了, 执行 pod install 下载库时,出现错误 解决 在 Podfile 加上 source ‘https://gith ...
- SQL如何使用快照恢复之前的数据
什么是快照 数据库快照是SQL server 2005的一个新功能.给出的定义如下 数据库快照是数据库的只读静态视图.在创建时每个数据库快照在事务上都与源数据库一致.在创建数据库快照时,源数据库通常会 ...
- Goldengate 部署oracle10g在 rac asm环境,完整教程
前言 Goldengate再rac 环境部署,和单机部署区别还是有点大,主要存在环境上. 环境 oracle10g ,sid=rac 准备工作 1.在rac节点,配置监听动态注册,确保goldenga ...
- web之请求转发与重定向
请求转发: 重定向: