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>
#include<string.h>
typedef struct{
  int id;
  int parent;
  int left;
  int right;
}Node;
void buildMaxHeap(Node *T,int n){
  int i;
  for(i=n/2;1<=i;i--){
    maxHeapify(T, i, n);
  }
}
int maxHeapify(Node *T,int i,int n){
  int l,r;
  int largest;
  Node A;
  
  l=2*i;
  r=2*i+1;
  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){
    A = T[i];
    T[i]=T[largest];
    T[largest]=A;
    maxHeapify(T, largest,n);
  }
}
int main(){
  int n,x;
  int i;
  Node *T;
  
  scanf("%d",&n);
  T=(Node*)malloc(sizeof(Node)*(n+1));
  for(i=1;i<=n;i++){//入力
    scanf("%d",&x);
    T[i].id=x;
    if(2*i<=n) T[2*i].parent=x;
    if(2*i+1<=n) T[2*i+1].parent=x;	
    if(i%2==0)T[i/2].left=x;
    else T[(i-1)/2].right=x;
  }
  buildMaxHeap(T,n);
  for(i=1;i<=n;i++){//出力
    printf(" %d",T[i]);
    }
  printf("\n"); 
  return 0;
}