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)

#include <stdio.h>

void MaxH(int *,int);

void BMax(int *);

int n;

int main()
{
  int i;

  scanf("%d",&n);

  int A[n+1];

  for(i=1;i<=n;i++)
    {
      scanf("%d",&A[i]);
    }

  BMax(A);

  for(i=1;i<=n;i++)printf(" %d",A[i]);


  printf("\n");
  
  return 0;
}

void MaxH(int *A,int i)
{
  int L,R,lg,koukan;

  L=2*i;
  R=2*i+1;

  if(L<=n&&A[L]>A[i])lg=L;
  else lg=i;

  if(R<=n&&A[R]>A[lg])lg=R;

  if(lg!=i)
    {
      koukan=A[i];
      A[i]=A[lg];
      A[lg]=koukan;

      MaxH(A,lg);
    }
}

void BMax(int *A)
{
  int i;
  for(i=n/2;i>=1;i--)
    {
      MaxH(A,i);
    }
}