Problem:
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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 2000000

int h[MAX+1],size = 0;

void maxHeapify(int i){
  int l,r,largest,ex;
  
  l = 2*i;
  r = 2*i+1;

  if(l <= size && h[l] > h[i])largest = l;
  else largest = i;

  if(r <= size && h[r] > h[largest])largest = r;
  
  if(largest != i){
    ex = h[i];
    h[i] = h[largest];
    h[largest] = ex;
    maxHeapify(largest);
  }
}

void IncreaseKey(int i,int k){
  int ex;
  
  h[i] = k;
  while(i > 1 && h[i/2] < h[i]){
    ex = h[i];
    h[i] = h[i/2];
    h[i/2] = ex;
    
    i /= 2;
  }

}

void Insert(int key){
  size++;
  h[size] = -1;
  IncreaseKey(size,key);
}

void extractMax(){
  if(size < 1)return;
  
  printf("%d\n",h[1]);
  h[1] = h[size];
  size--;
  
  maxHeapify(1);
}

int main(){
  int i,k;
  char a[7];
  
  for(i = 0;i < MAX; i++){
  scanf("%s",a);

  if(a[0] == 'i'){
    scanf("%d",&k);
    if(k < 0 || k > 2000000000)return 0;
    Insert(k);
  }
  else if(a[1] == 'x')extractMax();
  else return 0;

}
}