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>

int get_int() {
    int n = 0;
    int c = getchar_unlocked();
    if(c == 45) {
        c = getchar_unlocked();
        while(47 < c && c < 58) n = 10 * n + (c & 0xf), c = getchar_unlocked();
        return -n;
    } else if(c < 48 || 57 < c) return c;
    while(47 < c && c < 58) n = 10 * n + (c & 0xf), c = getchar_unlocked();
    return n;
}
void put_int(int n) {
    if(!n) {
        putchar_unlocked('0');
        return;
    }
    if(n < 0) n = -n, putchar_unlocked('-');
    char buf[11];
    int i = 0;
    while(n) buf[i++] = (char)(n % 10 + '0'), n /= 10;
    while(i--)putchar_unlocked(buf[i]);
}
void put_str(char *str) { while(*str) putchar_unlocked(*str++); }

void _maxheap(int *A, const int i, const int N) {
    int l = i * 2;
    int r = i * 2 + 1;
    int max = i;
    if(l <= N && A[l] > A[max]) max = l;
    if(r <= N && A[r] > A[max]) max = r;
    if(max != i) {
        int tmp = A[i];
        A[i] = A[max];
        A[max] = tmp;
        _maxheap(A, max, N);
    }
}
void maxheap(int *A, const int N) {
    for(int i = N / 2; i > 0; --i) _maxheap(A, i, N);
}
int main(int argc, char **argv) {
    int h = get_int();
    int a[h + 1];
    for(int i = 1; i <= h; ++i) a[i] = get_int();
    maxheap(a, h);
    for(int i = 1; i <= h; ++i) putchar_unlocked(' '), put_int(a[i]);
    putchar_unlocked('\n');
    return 0;
}