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::fmt::Write;

fn main() {
    let n: usize = read();
    let mut vec: Vec<i32> = read_as_vec();
    build_max_heap(n, &mut vec);
    println!(" {}", join(' ', &vec));
}

fn build_max_heap(len: usize, slice: &mut [i32]) {
    for i in (0..(len / 2)).rev() {
        max_heapify(len, slice, i);
    }
}

fn max_heapify(len: usize, slice: &mut [i32], idx: usize) {
    let left  = idx * 2 + 1;
    let right = idx * 2 + 2;
    let me = slice[idx];
    if right < len {      // 子ノードが2つ
        let max = if slice[left] > slice[right] { left } else { right };
        if slice[max] > me {
            slice[idx] = slice[max];
            slice[max] = me;
            max_heapify(len, slice, max);
        }
    } else if left < len { // 子ノードが1つ
        if slice[left] > me {
            slice[idx] = slice[left];
            slice[left] = me;
            max_heapify(len, slice, left);
        }

    }
}

fn read<T: std::str::FromStr>() -> T {
    let mut input = String::new();
    std::io::stdin().read_line(&mut input).unwrap();
    input.trim().parse::<T>().ok().unwrap()
}

fn read_as_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>()
        .split_whitespace()
        .map(|e| e.parse::<T>().ok().unwrap())
        .collect()
}

fn join<T: std::fmt::Display>(delimiter: char, arr: &[T]) -> String {
    let mut text = String::new();
    for e in arr.iter() {
        write!(text, "{}", e).unwrap();
        text.push(delimiter);
    }
    text.pop();
    text
}