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
#define INFTY 1
int H,A[MAX+1];
void maxHeapify(int i){
  int r , l , largest;
  int 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;
    return maxHeapify(largest);

  }
}


int extract(){
  int maxv;
  if(H < 1)return -INFTY;
  maxv = A[1];
  A[1] =A[H--];
  maxHeapify(1);
  return maxv;
}
void increaseKey(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]=-INFTY;
  increaseKey(H,key);
}
int main(){
  int key;
  char B[10];
  while(1){
    scanf("%s",B);
    if(B[0] == 'e' && B[1]=='n')break;
    if(B[0] == 'i'){
      scanf("%d",&key);
      insert(key);
    }
    else{
      printf("%d\n",extract());
    }

  }
  return 0;
}