Problem:
MathJax.Hub.Config({ tex2jax: { inlineMath: [["$","$"], ["\\(","\\)"]], processEscapes: true }});

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 Go (judge verdict: Accepted)

// AOJ ALDS1_9_B: http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=ALDS1_9_B

package main

import (
	"bufio"
	"fmt"
	"os"
	"strconv"
)

type maxHeap struct {
	heap []int
	size int
}

func nextInt(sc *bufio.Scanner) int {
	sc.Scan()
	v, _ := strconv.Atoi(sc.Text())
	return v
}

func (mh maxHeap) buildMaxHeap() {
	for i := mh.size / 2; i >= 1; i-- {
		mh.maxHeapify(i)
	}
}

func (mh maxHeap) maxHeapify(i int) {
	l := mh.left(i)
	r := mh.right(i)

	// 自分、左の子、右の子で値が最大のnodeを選ぶ
	largest := i
	if l <= mh.size && mh.heap[l] > mh.heap[i] {
		largest = l
	}

	if r <= mh.size && mh.heap[r] > mh.heap[largest] {
		largest = r
	}

	// iの子のほうが値が大きい場合
	if i != largest {
		mh.heap[i], mh.heap[largest] = mh.heap[largest], mh.heap[i]
		mh.maxHeapify(largest) // 再帰的に呼び出し
	}
}

func (mh maxHeap) left(i int) int {
	return 2 * i
}

func (mh maxHeap) right(i int) int {
	return 2*i + 1
}

func (mh maxHeap) printHeap() {
	buf := bufio.NewWriter(os.Stdout)
	for i := 1; i < mh.size+1; i++ {
		buf.Write([]byte(fmt.Sprintf(" %d", mh.heap[i])))
	}
	buf.Write([]byte("\n"))
	buf.Flush()
}

func main() {
	sc := bufio.NewScanner(os.Stdin)
	sc.Split(bufio.ScanWords)
	h := nextInt(sc)

	mh := maxHeap{make([]int, h+1), h}

	for i := 1; i < h+1; i++ {
		mh.heap[i] = nextInt(sc)
	}

	mh.buildMaxHeap()
	mh.printHeap()
}