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 <stdbool.h>
#include <stdint.h>
#include <stdlib.h>

#define UTIL_SWAP(x, y)                                                                                                                    \
  do {                                                                                                                                     \
    typeof(x) _a = x;                                                                                                                      \
    x = y;                                                                                                                                 \
    y = _a;                                                                                                                                \
  } while (0)

#define UTIL_MAX(x, y) ((x) < (y) ? (y) : (x))

static int64_t left_node_value(int64_t arr[], int32_t index, size_t len)
{
  if ((index * 2) <= len) {
    return arr[index * 2 - 1];
  } else {
    return INT64_MIN;
  }
}

static int64_t right_node_value(int64_t arr[], int32_t index, size_t len)
{
  if (((index * 2) + 1) <= len) {
    return arr[(index * 2) + 1 - 1];
  } else {
    return INT64_MIN;
  }
}

static void max_heapfy(int64_t arr[], size_t len, int32_t index)
{
  int64_t current = arr[index - 1];
  int64_t left = left_node_value(arr, index, len);
  int64_t right = right_node_value(arr, index, len);

  int32_t max = UTIL_MAX(current, UTIL_MAX(left, right));
  int32_t not_heapfy_index = index;
  if (max == current) {
    return;
  } else if (max == left) {
    not_heapfy_index = index * 2;
    UTIL_SWAP(arr[index - 1], arr[not_heapfy_index - 1]);
  } else {
    not_heapfy_index = index * 2 + 1;
    UTIL_SWAP(arr[index - 1], arr[not_heapfy_index - 1]);
  }
  max_heapfy(arr, len, not_heapfy_index);
}

void maximum_heap_build(int64_t arr[], size_t len)
{
  int32_t i;
  for (i = (len + 1) / 2; 0 < i; i--) {
    max_heapfy(arr, len, i);
  }
}

#if !defined(TEST)
#include <stdio.h>
int64_t buf[2 ^ 500000];

int32_t main(void)
{
  int32_t num;
  scanf("%d", &num);

  uint32_t i;
  for (i = 0; i < num; i++) {
    scanf("%ld", &buf[i]);
  }

  maximum_heap_build(buf, num);

  for (i = 0; i < num; i++) {
    printf(" %ld", buf[i]);
  }
  printf("\n");

  return 0;
}
#endif