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::{self, Read};
use std::str::FromStr;

pub struct Scanner<R: Read> {
    reader: R,
}
impl<R: Read> Scanner<R> {
    pub fn new(reader: R) -> Scanner<R> {
        Scanner { reader: reader }
    }
    pub fn read<T: FromStr>(&mut self) -> T {
        let s = self
            .reader
            .by_ref()
            .bytes()
            .map(|c| c.expect("failed to read char") as char)
            .skip_while(|c| c.is_whitespace())
            .take_while(|c| !c.is_whitespace())
            .collect::<String>();
        s.parse::<T>().ok().expect("failed to parse token")
    }
}

fn left(i: usize) -> usize {
    i*2
}

fn right(i: usize) -> usize {
    i*2+1
}

fn max_heapify(a: &mut Vec<i64>, i: usize, h: usize) {
    let l = left(i);
    let r = right(i);

    // 左の子、自分、右の子で最大のノードを選ぶ
    let mut largest = if l <= h && a[l] > a[i] {
        l
    } else {
        i
    };

    if r <= h && a[r] > a[largest] {
        largest = r;
    }

    if largest != i { // 自身より子の方が大きい場合
        a.swap(i, largest);
        max_heapify(a, largest, h);
    }
}

fn build_max_heap(a: &mut Vec<i64>, h: usize) {
    // h/2で子を持つノードの最後尾から順に計算する
    for i in (1..h/2+1).rev() {
        max_heapify(a, i, h);
    }
}

fn main() {
    let sin = io::stdin();
    let sin = sin.lock();
    let mut sc = Scanner::new(sin);

    let h: usize = sc.read();
    let mut vals: Vec<i64> = vec![0; h+1];

    for i in 1..h+1 {
        vals[i] = sc.read();
    }
    build_max_heap(&mut vals, h);

    println!(" {}", 
        vals[1..].iter()
        .map(|v| v.to_string())
        .collect::<Vec<String>>()
        .join(" "));
}