# 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：

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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