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  N 2000000
#define INF -2100000000

int h,a[N+1];

void maxHeapify(int i){
  int l,r,largest,temp;
  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){
    temp=a[i];
    a[i]=a[largest];
    a[largest]=temp;
    maxHeapify(largest);
  }
}

int extract(){
  int max;
  if(h<1) return INF;
  max=a[1];
  a[1]=a[h--];
  maxHeapify(1);
  return max;
}

void heapincreasekey(int i,int key){
  int temp;
  
  if(key<a[i]) return;
  a[i]=key;
  
  while(i>1&&a[i/2]<a[i]){
    temp=a[i];
    a[i]=a[i/2];
    a[i/2]=temp;
    i=i/2;
  }
}

void insert(int key){
  h++;
  a[h]=INF;
  heapincreasekey(h,key);
}

int main(){
  int key;
  char c[10];

  while(1){
    scanf("%s",c);
    if(c[0]=='e'&&c[1]=='n') break;
    if(c[0]=='i'){
      scanf("%d",&key);
      insert(key);
    }
    else{
      printf("%d\n",extract());
    }
  }
  
  return 0;
}