PAT_A1147#Heaps
Source:
Description:
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, orMin Heap
for a min heap, orNot 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
Keys:
Attention:
- 有点水了哈0,0;
Code:
/*
Data: 2019-06-29 16:15:43
Problem: PAT_A1147#Heaps
AC: 19:20 题目大意:
判断给定的完全二叉树是否是堆
输入:
第一行给出测试数M(<=100)和结点数N[1,1e3]
接下来N行,逐层给出完全二叉树的各个键值
输出:
大根堆,小根堆,非堆;
接下来输出二叉树的后序遍历 基本思路:
静态树存储BST,遍历判断是否为堆
*/
#include<cstdio>
#include<vector>
using namespace std;
const int M=1e3+;
int bst[M],Min,Max,n,m;
vector<int> path; void Travel(int root)
{
if(root > n)
return;
if(root!=)
{
if(bst[root/] > bst[root])
Min=;
if(bst[root/] < bst[root])
Max=;
}
Travel(root*);
Travel(root*+);
path.push_back(bst[root]);
} int main()
{
#ifdef ONLINE_JUDGE
#else
freopen("Test.txt", "r", stdin);
#endif // ONLINE_JUDGE scanf("%d%d", &m,&n);
while(m--)
{
for(int i=; i<=n; i++)
scanf("%d", &bst[i]);
Min=;
Max=;
path.clear();
Travel();
if(Max) printf("Max Heap\n");
else if(Min) printf("Min Heap\n");
else printf("Not Heap\n");
for(int i=; i<n; i++)
printf("%d%c", path[i],i==n-?'\n':' ');
} return ;
}
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