Problem:
MathJax.Hub.Config({ tex2jax: { inlineMath: [["$","$"], ["\\(","\\)"]], processEscapes: true }});

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: Time Limit Exceeded)

#include <stdio.h>
#include <stdlib.h>
typedef struct Tree{
  int key;
  int parent;
  int left;
  int right;
}Tree;

int main(){
  int H,i,j,store;
  Tree *T;
  scanf("%d",&H);
  T=(Tree*)malloc(sizeof(Tree)*H);
  for(i=0;i<H;i++){
    T[i].key=2000000001;
    T[i].parent=-1;
    T[i].left=-1;
    T[i].right=-1;
  }
  
  for(i=0;i<H;i++){
    scanf("%d",&T[i].key);
    T[i].parent=(i+1)/2-1;
    if(2*i+1<H){
      T[i].left=2*i+1;
      if(2*i+2<H)
	T[i].right=2*i+2;
    }
  }
  
  for(i=T[H-1].parent;i!=-1;i--){
    j=i;
    while(j<H){
      if(T[T[j].left].key>=T[T[j].right].key){
	if(T[T[j].left].key>T[j].key){
	  store=T[j].key;
	  T[j].key=T[T[j].left].key;
	  T[T[j].left].key=store;
	  j=T[j].left;
	}
	else
	  j=H;
      }
      else
	if(T[T[j].right].key>T[j].key){
	  store=T[j].key;
	  T[j].key=T[T[j].right].key;
	  T[T[j].right].key=store;
	  j=T[j].right;
	}
	else
	  j=H;
    }
  }

  
  for(i=0;i<H;i++){
    printf(" %d",T[i].key);
  }
  printf("\n");
  return 0;
}