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>
#include<string.h>
#define MAX 2000000
#define INFTY (1<<30)

int H,A[MAX+1];

void maxHeapify(int i){
  int l,r,largest,t;
  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){
    t=A[i];
    A[i]=A[largest];
    A[largest]=t;
    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 k;
  if(key < A[i]) return;
  A[i] = key;
  while(i>1 && A[i/2]<A[i]){
    k=A[i];
    A[i]=A[i/2];
    A[i/2]=k;
    i=i/2;
  }
}

void insert(int key){
  H++;
  A[H] = -INFTY;
  increaseKey(H,key);
}

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

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

  return 0;
}