1147 Heaps (30 分)_cmd heap-优快云博客

本文链接：https://blog.youkuaiyun.com/zxc1364787928/article/details/94446613

本博客介绍了一个算法，用于判断给定的完全二叉树是否满足最大堆或最小堆的性质，并通过后序遍历输出结果。示例输入包含多个测试用例，每个用例包括树中节点的数量和节点的值，输出则指示树是否为最大堆、最小堆或根本不是堆。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

1147 Heaps (30 分)

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

===================================================

1155题是这题的升级版，所以要重视近些年的真题，很有可能改动一下继续考。

===================================================

#include <iostream>
using namespace std;
int m, n, isMax, isMin, a[1005] = { 0 };
void postorder(int index){
	if(index > n) return;
	postorder(index * 2);
	postorder(index * 2 + 1);
	printf("%d%s", a[index], index == 1 ? "\n" : " ");
}
int main(){
	cin >> m >> n;
	while(m--){
		isMin = isMax = 1;
		for(int i = 1; i <= n; i++)
			scanf("%d", &a[i]);
		for(int i = 2; i <= n; i++){
			if(a[i] < a[i / 2]) isMin = 0;
			if(a[i] > a[i / 2]) isMax = 0;
		}
		if(isMin)
			printf("Min Heap\n");
		else
			printf("%s\n", isMax == 0 ? "Not Heap" : "Max Heap");
		postorder(1);
	}
	return 0;
}