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>

void heap(int a[],int i,int n){
    int l,r,max,k;
    if(i*2+1>=n){
        return;
    }else if(i*2+2==n){
        if(a[i]<a[n-1]){
            r=a[i];
            a[i]=a[n-1];
            a[n-1]=r;
        }
        return;
    }else{
        l = i*2+1;
        r = i*2+2;
        max = i;
        if(a[i]<a[l]){
            max = l;
        }
        if(a[max]<a[r]){
            max = r;
        }
        if(max!=i){
            k=a[i];
            a[i]=a[max];
            a[max]=k;
            heap(a,max,n);
        }
    }
}

int main(void){
    int n,i,j,k;
    int *a;
    scanf("%d",&n);
    a = (int*)malloc(sizeof(int)*n);
    for(i=0;i<n;i++){
        scanf("%d",&a[i]);
    }
    for(i=n/2;i>=0;i--){
        heap(a,i,n);
    }
    for(i=0;i<n;i++){
        printf(" %d",a[i]);
    }
    printf("\n");
    free(a);
    return 0;
}