PAT (Advanced Level)-1147 Heaps

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

 

思路：完全二叉的树的最大堆和最小堆的 不需要重新维护堆 只需判断 后序遍历递归输出

完全二叉树的性质：最后一个非叶子结点是第n/2个结点 （堆顶以1开始）

#include <iostream>
#include <vector>
using namespace std;


int n,h[1001];
vector <int>v;
void postorder(int i)
{
	if (i <= n)
	{
		postorder(i * 2);
		postorder(i * 2+1);
		v.push_back(h[i]);
	}

}
int max_h()
{
	int i;
	for (i = n / 2; i >= 1; i--)
	{
		int t;
		if (h[i * 2] > h[i])t = i * 2;
		else t = i;
		if (i * 2 + 1 == n)
		{
			if (h[i * 2 + 1] > h[t])t = i * 2 + 1;
		}
		if (t != i)return 0;
	}
	return 1;
}

int min_h()
{
	int i;
	for (i = n / 2; i >= 1; i--)
	{
		int t;
		if (h[i * 2] < h[i])t = i * 2;
		else t = i;
		if (i * 2 + 1 == n)
		{
			if (h[i * 2 + 1] < h[t])t = i * 2 + 1;
		}
		if (t != i)return 0;
	}
	return 1;
}
int main()
{
	int i,m;
	cin >>m>> n;
	for (int j = 0; j < m; j++)
	{
		int flag = 1;
		for (i = 1; i <= n; i++)
		{
			scanf("%d", &h[i]);
		}
		
		if (max_h())cout << "Max Heap" << endl;
		else if(min_h())cout << "Min Heap" << endl;
		else cout << "Not Heap" << endl;

		postorder(1);
		for (i = 0; i < v.size(); i++)
		{
			if (i == 0)printf("%d", v[i]);
			else printf(" %d", v[i]);
		}
		printf("\n");
		v.clear();
	}
}