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>

int Parent(int );
int Left(int );
int Right(int );

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

int H; //heap-size = n
int main() {
  int n;
  int i;
  int j;
  int *array;

  scanf("%d",&n);
  H = n;
  array = (int *)malloc(n * sizeof(int));
  for(i=0;i<n;i++) scanf(" %d",&array[i]);

  buildMaxHeap(array);

  for(i=0;i<n;i++) printf(" %d",array[i]);
  printf("\n");
  return 0;
}

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

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

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

void maxHeapify(int *A,int i) {  //i->+1
  int l,r;
  int largest;
  int tmp;
  l = Left(i); //l->+1
  r = Right(i); //r->+1
  // 左の子、自分、右の子で値が最大のノードを選ぶ
  if(l > H && r > H) largest = i;
  else {
    if(l <= H && A[l-1] > A[i-1]) largest = l;  //largest->+1
    else largest = i; //largest->+1
    if(r <= H)
      if(r <= H && A[r-1] > A[largest-1]) largest = r; //largest->+1
  }

  //printf("checklargest %d:%d\n",largest,A[largest-1]);

  //printf("%d : i->%d, left->%d, right->%d\n",i,A[i-1],A[l-1],A[r-1]);
  if(largest != i) {
    tmp = A[i-1];
    A[i-1] = A[largest-1];
    A[largest-1] = tmp;

    maxHeapify(A, largest); //
  }
  
}

void buildMaxHeap(int *A) {
  for(int i=H/2;i>0;i--) maxHeapify(A,i);
}