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>
#define mymalloc(p,n) {p = malloc((n)*sizeof(*p));if ((p)==NULL) {printf("not enough memory?\n"); exit(1);};}

typedef struct{
	long *A;
	long n;
}heap;

int p(int i){return (i/2);}
int l(int i){return (2*i);}
int r(int i){return (2*i+1);}

void maxHeapify(heap *H,long i){
	long *A;
	long n,largest,temp;
	A=H->A;n=H->n;
	if(l(i)<=n && A[l(i)] > A[i]){
		largest = l(i);
	}else{
		largest = i;
	}
	if(r(i)<=n && A[r(i)] > A[largest]){
		largest = r(i);
	}
	if(largest!=i){
		temp=A[i];A[i]=A[largest];A[largest]=temp;
		maxHeapify(H,largest);
	}
}

void buildMaxHeap(heap *H){
	long i;
	for(i=H->n/2;i>=1;i--){
		maxHeapify(H,i);
	}
}

main(){
	long n,i,m;
	heap *H;
	scanf("%ld",&n);
	mymalloc(H,1);
	H->n=n;
	mymalloc(H->A,n+1);
	for(i=1;i<=n;i++){
		scanf(" %ld",&m);
		H->A[i]=m;
	}
	buildMaxHeap(H);
	for(i=1;i<=n;i++){
		printf(" %ld",H->A[i]);
	}
	printf("\n");
	return 0;
}