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>
#define N 2000001

void buildMaxHeap(int* a_p, int h);
void maxh(int* a_p, int ix, int h);
int parent(int i);
int left(int i);
int right(int i);
void swap(int *a_p, int *b_p);
void insert(int* a_p, int key, int* s_p);
void inc(int* a_p, int ix, int key);
int exmax(int* a_p, int* s_p);

int main(){
  int n, i, j;
  int ix;
  int key;
  int size = 1;
  char cmd[30];
  int a[N]; // 1-origin

  while(1){
    scanf("%s", cmd);
    if( cmd[0] == 'i'){
      scanf("%d" ,&key);
      insert( a, key, &size ); 
    }else if( cmd[0] == 'e' && cmd[1] == 'x' ){
      printf("%d\n", exmax( a, &size) );
    }else{ // end
      break;
    }  
  }

  return 0; // Don't forget return 0
}

void insert(int* a_p, int key, int* s_p){
  (*s_p)++;
  a_p[*s_p] = -1;
  inc( a_p, *s_p, key); 
}

void inc(int* a_p, int ix, int key){
  a_p[ix] =key;
  while( ix>1 && a_p[parent(ix)] < a_p[ix] ){
    swap( &(a_p[ix]), &(a_p[parent(ix)]) );
    ix = parent(ix);
  }
}

int exmax(int* a_p, int* s_p){
  int max = a_p[1];
  a_p[1] = a_p[*s_p];
  (*s_p)--;
  maxh( a_p, 1 ,*s_p);
  return max;
}

void buildMaxHeap(int* a_p, int h){
  int i;
  for( i=h/2 ; i>=1 ; i--){
    maxh( a_p ,i ,h );
  }
}

void maxh(int* a_p, int ix, int h){
  int largest;
  int l = left(ix);
  int r = right(ix);

  if( l<=h && a_p[l]>a_p[ix] ){
    largest = l;
  }else {
    largest = ix;
  }
  if( r <= h && a_p[r]>a_p[largest] ){
    largest = r;
  }

  if(largest!=ix){
    swap( &(a_p[ix]) ,&(a_p[largest]) );
    maxh( a_p ,largest ,h);
  }

}

int parent(int i){
  return floor(i/2);
}

int left(int i){
  return 2*i;
}

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

void swap(int *a_p, int *b_p){
  int k;
  k    = *a_p;
  *a_p = *b_p;
  *b_p = k;
}