PAT (Advanced Level) 1147 Heaps (30 分)

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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

Code:

#include <iostream>
#include <vector>
#pragma warning(disable:4996)
using namespace std;

int m, n, cnt = 0;

void postOrder(vector<int>& vec, int root)
{
	if (root > n)
		return;
	postOrder(vec, root * 2);
	postOrder(vec, root * 2 + 1);
	printf("%d%c", vec[root], (++cnt == n ? '\n' : ' '));
}

int main()
{

	scanf("%d%d", &m, &n);
	for (int q = 0; q < m; q++)
	{
		vector<int> vec(n + 1);
		for (int i = 1; i < vec.size(); i++)
			scanf("%d", &vec[i]);
		bool maxhp = true, minhp = true;
		for (int i = 1; i <= n / 2; i++)
		{
			int lchild = 2 * i, rchild = 2 * i + 1;
			if (lchild <= n && vec[i] < vec[lchild]) {
				maxhp = false; break;
			}
			if (rchild <= n && vec[i] < vec[rchild]) {
				maxhp = false; break;
			}
		}
		if (maxhp) {
			printf("Max Heap\n");
		}
		else {
			for (int i = 1; i <= n / 2; i++)
			{
				int lchild = 2 * i, rchild = 2 * i + 1;
				if (lchild <= n && vec[i] > vec[lchild]) {
					minhp = false; break;
				}
				if (rchild <= n && vec[i] > vec[rchild]) {
					minhp = false; break;
				}
			}
			if (minhp)
				printf("Min Heap\n");
			else
				printf("Not Heap\n");
		}
		postOrder(vec, 1);
		cnt = 0;
	}
	return 0;
}

思路:

考完全二叉树的知识,如果一棵完全二叉树有n个节点,那么:

  1. 他要被储存在[1, n]而不是[0, n-1]的数组中;
  2. 如果根节点的序号是i,那么左子树是2i,右子树是2i+1;
  3. 非叶节点的序号范围是[1, n/2]。

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