PAT (Advanced Level) Practice 1147 Heaps (30 分) 凌宸1642

2024-01-24 17:24:46

PAT (Advanced Level) Practice 1147 Heaps (30 分) 凌宸1642

题目描述：

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.

译：在计算机科学中，堆是一种特殊的基于树的数据结构，它满足堆属性：如果 P 是 C 的父节点，则 P 的键（值）要么大于或等于（在最大堆中） ) 或小于或等于（在最小堆中）C 的键。堆的常见实现是二叉堆，其中树是完全二叉树。（引自* https://en.wikipedia.org/wiki/Heap_(data_structure)）

你的工作是判断一个给定的完整二叉树是否是一个堆。

Input Specification (输入说明):

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.

译：每个输入文件包含一个测试用例。对于每种情况，第一行给出两个正整数：M（≤ 100），要测试的树的数量；和 N (1 < N ≤ 1,000)，分别是每棵树中的键数。然后是 M 行，每行包含 N 个不同的整数键（都在 int 范围内），这给出了完整二叉树的层序遍历序列。

output Specification (输出说明):

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.

译：对于每个给定的树，如果它是最大堆，则在一行中打印 Max Heap，如果是最小堆，则打印为Min Heap，如果根本不是堆，则打印为 Not Heap。然后在下一行打印树的后序遍历序列。所有数字之间用空格隔开，行首或行尾不得有多余空格。

Sample Input (样例输入):

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 (样例输出):

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

The Idea：

直接建树的话，最后两个测试点会超时。
利用堆的性质，可以不见树即可判断是否是堆，然后由于需后序遍历，最后一个结点一定是根节点，所以可以直接对数组使用 dfs 。

The Codes:

#include<bits/stdc++.h>
using namespace std ;
int layer[1010] , m , n ;
void dfs(int root){
	if(root > n ) return ;
	dfs(root * 2) ;
	dfs(root * 2 + 1) ;
	cout << layer[root] << ((root == 1)?'\n':' ') ;
 }
int main(){
	cin >> m >> n ;
	for(int i = 0 ; i < m ; i ++){
		for(int j = 1 ; j <= n ; j ++) cin >> layer[j] ;
		bool minHeap = true , maxHeap = true ;
		for(int j = 1 ; j <= n ; j ++){
			if(j > 1 && layer[j/2] > layer[j]) minHeap = false ;
			if(j > 1 && layer[j/2] < layer[j]) maxHeap = false ;
		}
		if(maxHeap) cout << "Max Heap" << endl ;
		else if(minHeap) cout << "Min Heap" << endl ;
		else cout << "Not Heap" << endl ;
		dfs(1) ;
	}
	return 0 ;
}

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