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 INFTY (1<<30)

void insert(int);
int extractmax(void);
void maxHeap(int);
void increase(int,int);
int H,A[N+1];

int main()
{
  
  char c[10];
  int n;
  while(1){
    scanf("%s",c);

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

void increase(int i,int n)
{
  int x;
  if(n<A[i])return;
  A[i]=n;
  while(i>1 && A[i/2]<A[i])
    {
      x=A[i];
      A[i]=A[i/2];
      A[i/2]=x;
      i=i/2;
    }
}

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

int extractmax(void)
{
  int max;
  if(H<1)return -INFTY;
  max=A[1];
  A[1]=A[H--];
  maxHeap(1);
  return max;
}

void maxHeap(int i)
{
  int l,r,largest,x;
  l = i*2;
  r = i*2+1;
  /* 左の子、自分、右の子で値が最大のノードを選ぶ*/
if (l<=H && A[l]>A[i])
  largest = l;
else
  largest = i;

if(r<=H && A[r] > A[largest])
  largest = r;
 if (largest != i)
   {x=A[i];
    A[i]=A[largest];
    A[largest]=x;

   maxHeap(largest);/* 再帰的に呼び出し*/ }
}