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 <math.h>

void heapInsert(int);
//void heapIncreasekey(int, int);
void heapExtractMax();
void maxHeapify(int);

int parent(int i){ return i/2;}
int left(int i){ return i*2;}
int right(int i){ return i*2+1;}

int a[2000000];
int key;
int size =1;

int main(){

  char o[10];

  while(1){

    scanf("%s", o);

    if(o[0]=='e' && o[1]=='n') break;

    if(o[0]=='i'){
      scanf("%d", &key);
      heapInsert(key);     
    }
    if(o[0]=='e' && o[1]=='x'){
      heapExtractMax();
    }
  }
  return 0;
}
void heapInsert(int key){

  int i, temp;
  size++;
  
  a[size]=key;

  i=size;
  while(1<i && a[parent(i)]< a[i] ){
    temp=a[parent(i)];
    a[parent(i)]=a[i];
    a[i]=temp;
    i=parent(i);
  }
}
void heapExtractMax(){

  int max;
  
  if(size<1) printf("error\n");

  printf("%d\n", a[1]);
  max=a[1];
  a[1]=a[size];
  size--;
  maxHeapify(1);
}
void maxHeapify(int i){ //error

  int left =(i*2);
  int right=(i*2+1);
  int largest, temp;
  
  //左の子、自分、右の子でMAXを選ぶ

  if(left<=size && a[i]< a[left]){ //左の子存在、左の子より小さい
    largest=left; //あくまで節点の場所
  }else largest=i;

  if(right<=size && a[largest]<a[right]){

    largest=right;
  }

  if(largest != i){

    temp=a[largest];

    a[largest]=a[i];
    a[i]=temp;

    maxHeapify(largest);
  }
}