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: Runtime Error)

#include<stdio.h>

#define Max 500001

typedef struct{
  int key;
  int parent;
  int left;
  int right;
}Node;

void maxHeapify(Node *, int, int);
void buildMaxHeap(Node *, int);

int main(){
  int i, H;
  Node T[Max];

  scanf("%d",&H);
  for(i = 1; i <= H; i++){
    T[i].parent = NULL;
    T[i].left = NULL;
    T[i].right = NULL;
  }
  for(i = 1; i <= H; i++){
    scanf("%d",&T[i].key);
    if(i > 1) T[i].parent = T[i / 2].key;
  }
  for(i = 1; i <= H; i++){
    T[i].left = T[2 * i].key;
    if(2 * i > H) T[i].left = NULL;
    T[i].right = T[2 * i + 1].key;
    if(2 * i + 1 > H) T[i].right = NULL;
  }

  buildMaxHeap(T, H);
  
  for(i = 1; i < H; i++){
    printf(" %d", T[i].key);
  }
  printf(" %d\n", T[i].key);

  return 0;
}


void maxHeapify(Node *A, int i, int H){
  int l, r, largest, n;

  l = i * 2;
  r = i * 2 + 1;

  if((l <= H) && (A[l].key > A[i].key)) largest = l;
  else largest = i;

  if((r <= H) && (A[r].key > A[largest].key)) largest = r;

  if(largest != i){
    n = A[i].key;
    A[i].key = A[largest].key;
    A[largest].key = n;
    maxHeapify(A, largest, H);
  }

}

void buildMaxHeap(Node *A, int H){
  int i;

  for(i = H / 2; i >= 1; i--){
    maxHeapify(A, i, H);
  }
}