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 Rust (judge verdict: Accepted)

use std::io::*;
use std::str::FromStr;

fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed to parse token")
}

fn left(node_id: usize) -> usize {
    return node_id * 2;
}

fn right(node_id: usize) -> usize {
    return node_id * 2 + 1;
}

fn max_heapify(a: &mut Vec<i64>, node_id: usize) {
    let left = left(node_id);
    let right = right(node_id);
    let mut largest: usize;
    if left <= (a.len() - 1) && a[left] > a[node_id] {
        largest = left;
    } else {
        largest = node_id;
    }
    if right <= (a.len() - 1) && a[right] > a[largest] {
        largest = right;
    }

    if largest != node_id {
        a.swap(node_id, largest);
        max_heapify(a, largest);
    }
}

fn build_max_heap(a: &mut Vec<i64>) {
    for i in (1..(((a.len() - 1) / 2) + 1)).rev() {
        max_heapify(a, i);
    }
}

fn main() {
    let h: usize = read();
    let mut a: Vec<i64> = vec![0];
    for _ in 0..h {
        let number: i64 = read();
        a.push(number);
    }
    build_max_heap(&mut a);
    for (index, value) in a.iter().enumerate() {
        if index == 0 {
            continue;
        } else {
            print!(" {}", value);
        }
    }
    println!();
}