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: Runtime Error)

#include<stdio.h>
#include<stdlib.h>
#define tree_t long int*
#define first 0
int count; //??????heap???????????????

typedef struct node
{
    tree_t d;
    int number;
    int max_num;

} heap;


void swap(long int *n1,long int *n2)
{
    long int setp;
    count++;    // ????????????????¨?????¬????
    setp = *n1;
    *n1 = *n2;
    *n2 = setp;

}


void creat_heap(heap* in,int max_)//?°±????????????
{
    in->d = (tree_t)malloc(max_*sizeof(int));
    in->max_num = max_;
    in->number = -1;
}

void heap_down(heap* answer,int root,int botten)  //heap_down ??????
{
    int lchild = root*2+1;
    int rchild = root*2+2;
    int max_child;

    if(lchild < botten && rchild<botten)
    {
        if(answer->d[lchild]>=answer->d[rchild])
        {
            max_child = lchild;
        }
        else
        {
            max_child = rchild;
        }

        if(answer->d[root] < answer->d[max_child])
        {
            swap(&(answer->d[root]),(&answer->d[max_child]));
        }
        heap_down(answer,max_child,botten);
    }
    else if(lchild<botten){
        if(answer->d[root] < answer->d[lchild])
        {
            swap(&(answer->d[root]),(&answer->d[lchild]));
        }
        heap_down(answer,lchild,botten);
    }
    return;
}

void heap_up(heap* answer,int botten)   //heap_up??????(??¶?????¨?????°
{
    int root = 0;
    int parent;
    if(botten > root)
    {
        parent = (botten-1)/2;

        if(answer->d[parent] < answer->d[botten])
        {
            //printf("%d %d\n",answer->d[parent] , answer->d[botten]);
            swap(&(answer->d[parent]), &(answer->d[botten]));
            //printf("%d %d\n\n",answer->d[parent] , answer->d[botten]);
            heap_up(answer,parent);
        }
    }
    return;

}

void out_h(heap* answer)
{
    int i;
    for(i=0; i <= answer->number; i++)
    {
        printf(" %ld",answer->d[i]);
    }
    printf("\n");

}

int main()
{

    int max_;

    while(scanf("%d",&max_)!=EOF)
    {

        heap answer;
        creat_heap(&answer,max_);
        long int input;
        int i;
        for(i=0; i<max_; i++)
        {
            scanf("%ld",&input);
            answer.d[i] = input;
            answer.number++;
        }
        for(i = max_-1;i>=0;i--)
        {
            heap_down(&answer,i,max_);
        }

        out_h(&answer);



    }


    return 0;
}