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 INF 1000000
#define N 2000000
 
void insert(int *, int);
void increase(int *, int, int);
int extractmax(int *);
void maxHeapify(int *, int);
 
int n=0;
int S[N];
 
int main()
{
 
  int i=0, k, j, max[N];
  char com[10];
 
 
  while(1){
    scanf("%s", com);
     
    if(com[0] == 'i'){
      scanf("%d", &k);
      insert(S, k);
    }
 
    else if(com[1] == 'x'){
      max[i] = extractmax(S);
      i++;
    }
 
    else if(com[1] == 'n') break;
 
  }
 
  for(j = 0; j < i; j++){
    printf("%d\n", max[j]);
  }
 
  return 0;
 
}
 
void insert(int *S, int k)
{
  n += 1; 
  S[n] = -INF;
  increase(S, n, k);
}
 
void increase(int *S, int i, int key)
{
 
  int temp;
 
  S[i] = key;
 
  while(i > 1 && S[i/2] < S[i]){
    temp = S[i];
    S[i] = S[i/2];
    S[i/2] = temp;
    i = i/2;
  }
}
 
int extractmax(int *S)
{
  int max;
 
  max = S[1];
  S[1] = S[n];
  n--;
  maxHeapify(S, 1);
 
  return max;
}
 
void maxHeapify(int *A, int i)
{
 
  int l, r, largest;
  int temp;
 
  l = 2*i;
  r = 2*i+1;
   
  if(l <= n && A[l] > A[i]) largest = l;
  else largest = i;
 
  if(r <= n && A[r] > A[largest]) largest = r;
 
  if(largest != i){
    temp = A[i];
    A[i] = A[largest];
    A[largest] = temp;
    maxHeapify(A, largest);
  }
}