Problem:
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Maximum Heap

A binary heap which satisfies max-heap property is called max-heap. In a max-heap, for every node $i$ other than the root, $A[i] \leq A[parent(i)]$, that is, the value of a node is at most the value of its parent. The largest element in a max-heap is stored at the root, and the subtree rooted at a node contains values no larger than that contained at the node itself.

Here is an example of a max-heap.

Write a program which reads an array and constructs a max-heap from the array based on the following pseudo code.

$maxHeapify(A, i)$ move the value of $A[i]$ down to leaves to make a sub-tree of node $i$ a max-heap. Here, $H$ is the size of the heap.

1 maxHeapify(A, i)
2 l = left(i)
3 r = right(i)
4 // select the node which has the maximum value
5 if l ≤ H and A[l] > A[i]
6 largest = l
7 else
8 largest = i
9 if r ≤ H and A[r] > A[largest]
10 largest = r
11
12 if largest ≠ i　// value of children is larger than that of i
13 swap A[i] and A[largest]
14 maxHeapify(A, largest) // call recursively

The following procedure buildMaxHeap(A) makes $A$ a max-heap by performing maxHeapify in a bottom-up manner.

1 buildMaxHeap(A)
2 for i = H/2 downto 1
3 maxHeapify(A, i)

Input

In the first line, an integer $H$ is given. In the second line, $H$ integers which represent elements in the binary heap are given in order of node id (from $1$ to $H$).

Output

Print values of nodes in the max-heap in order of their id (from $1$ to $H$). Print a single space character before each value.

Constraint

$1 \leq H \leq 500,000$

$-2,000,000,000 \leq$ value of a node $\leq 2,000,000,000$

Sample Input 1

10
4 1 3 2 16 9 10 14 8 7

Sample Output 1

16 14 10 8 7 9 3 2 4 1

Reference

Introduction to Algorithms, Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, and Clifford Stein. The MIT Press.

Solution in C# (judge verdict: Accepted)

using System;
using System.Linq;

class Node{
  public int key = 0;
  public int parent = 0;
  public int left = 0;
  public int right = 0;
}

class Max{
  public void maxHeap(ref Node[] node,int i){
    int left = i * 2, right = i * 2 + 1,large = i;
    Node change = new Node();
    if(left < node.Length && node[left].key > node[i].key){
      large = left;
    }if(right < node.Length && node[right].key > node[large].key){
      large = right;
    }if(large != i){
      change = node[large];
      node[large] = node[i];
      node[i] = change;
      maxHeap(ref node,large);
    }
  }
}

class B9{
  public static void Main(){
   int n = int.Parse(Console.ReadLine());
   Node[] node = new Node[n + 1];
   for(int i = 0;i <= n;i++){
     node[i] = new Node();
   }
   int[] input = Array.ConvertAll(Console.ReadLine().Split(' '),int.Parse);
   for(int i = 0;i < n;i++){
     node[i + 1].key = input[i];
   }
   for(int i = 1;i <= n;i++){
     if(i / 2 > 0){
       node[i].parent = node[i / 2].key;
     }
     if(i * 2 <= n){
       node[i].left = i * 2;
     }
     if(i * 2 + 1 <= n){
       node[i].right = 2 * i + 1;
     }
   }
   Max max = new Max();
   for(int i = n / 2;i > 0;i--){
     max.maxHeap(ref node,i);
   }
   for(int i = 1;i <= n;i++){
     Console.Write(" " + node[i].key);
   }
   Console.WriteLine();
  }
}