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>
#include<stdlib.h>

struct node{
  int right;
  int left;
  int parent;
  int id;
};

#define MAX 500001
#define N 2000000001

struct node T[MAX];

int n;

int parent(int i){
  return i/2;
}

int left(int i){
  return 2*i;
}

int right(int i){
  return 2*i+1;
}

void insert(){
  int i;
  for(i = 1; i <= n/2; i++){
    T[2*i].parent = parent(i);
    T[2*i].left = left(i);
    T[2*i].right = right(i);
  }
}

void maxHeapify(int i){
  int h, l, r, largest;
  l = left(i);
  r = right(i);
  if(l <= n && T[l].id > T[i].id) largest = l;
  else largest = i;
  if(r <= n && T[r].id > T[largest].id) largest = r;
  if(largest != i){
    h = T[i].id;
    T[i].id = T[largest].id;
    T[largest].id = h;
    maxHeapify(largest);
  }
}

void buildMaxHeap(){
  int i;
  for(i=n/2; i>0; i--) maxHeapify(i);
}

void print(){
  int i;
  buildMaxHeap();
  for(i=1; i<=n; i++) printf(" %d", T[i].id);
}

int main(){
  int i;

  scanf("%d", &n);
  for( i = 1; i <= n; i++ ) scanf("%d", &T[i].id);
  insert();
  print();
  printf("\n");
  return 0;
}