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

Priority Queue

A priority queue is a data structure which maintains a set $S$ of elements, each of with an associated value (key), and supports the following operations:

$insert(S, k)$: insert an element $k$ into the set $S$

$extractMax(S)$: remove and return the element of $S$ with the largest key

Write a program which performs the $insert(S, k)$ and $extractMax(S)$ operations to a priority queue $S$.
The priority queue manages a set of integers, which are also keys for the priority.

Input

Multiple operations to the priority queue $S$ are given. Each operation is given by "insert $k$", "extract" or "end" in a line. Here, $k$ represents an integer element to be inserted to the priority queue.

The input ends with "end" operation.

Output

For each "extract" operation, print the element extracted from the priority queue $S$ in a line.

Constraints

The number of operations $\leq 2,000,000$

$0 \leq k \leq 2,000,000,000$

Sample Input 1

insert 8
insert 2
extract
insert 10
extract
insert 11
extract
extract
end

Sample Output 1

8
10
11
2

Solution in C (judge verdict: Accepted)

#include <stdio.h>
 
#define MAX 2000001
 
void insert(int);
void heapIncreaseKey(int, int);
int extract(void);
void maxHeapify(int);
 
int H=0, A[MAX];
 
int main()
{
  int key, Out[MAX], i, j;
  char input[10];
 
  while( 1 ){
    scanf("%s", input);
    if( input[0] == 'e' && input[1] == 'n' ) break;
    if( input[0] == 'i' ){
      scanf("%d", &key);
      insert(key);
    }
    else {
      Out[j] = extract();
      j++;
    }
  }
 
  for( i = 0 ; i < j ; i++ ) printf("%d\n", Out[i]);
   
  return 0;
}
 
void insert(int key)
{
  H++;
  A[H] = -1000000000;
  heapIncreaseKey(H, key);
}
 
void heapIncreaseKey(int i, int key)
{
  int tmp;
   
  if( key < A[i] ) return;
  A[i] = key;
  while( i > 1 && A[i/2] < A[i] ){
    tmp = A[i];
    A[i] = A[i/2];
    A[i/2] = tmp;
    i = i / 2;
  }
}
 
int extract()
{
  int max;
 
  if( H < 1 ) return -1000000000;
  max = A[1];
  A[1] = A[H--];
  maxHeapify(1);
  return max;
}
 
void maxHeapify(int i)
{
  int l, r, largest, tmp;
   
  l = 2 * i;
  r = 2 * i + 1;
   
  if( l <= H && A[l] > A[i] ) largest = l;
  else largest = i;
  if( r <= H && A[r] > A[largest] ) largest = r;
   
  if( largest != i ){
    tmp = A[i];
    A[i] = A[largest];
    A[largest] = tmp;
    maxHeapify(largest);
  }
}