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 1000000

int H, A[MAX + 1];

void maxHeapify(int);
void insert(int);
void increasekey(int, int);
int maxExtract(void);

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

  while(1){
    scanf("%s", &array[0]);
    if(!strcmp(array, "end")) break;
    if(!strcmp(array, "insert")){
      scanf("%d", &key);
      insert(key);
    }
    else printf("%d\n", maxExtract());
  }
  
  return 0;
}

void maxHeapify(int i){
  int r, l, largest, temp;

  l = i * 2;
  r = i * 2 + 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){
    temp = A[largest];
    A[largest] = A[i];
    A[i] = temp;
    maxHeapify(largest);
  }
}
void insert(int key){
  H++;  //H = H + 1 で木の一番深いレベルの一番右側の葉に入れる添え字を作る。
  A[H] = key;  
  increasekey(H, key);
}
void increasekey(int i, int key){
  int temp;
  
  while(i > 1 && A[i / 2] < A[i]){  //親と子を比較して、入れ替え。あるいはそのまま。
    temp = A[i / 2];
    A[i / 2] = A[i];
    A[i] = temp;
    i = i / 2;
  }
}
int maxExtract(void){
  int max; //最大値を extract したいから、木構造の root が最大値である。もちろん return するのは max。

  max = A[1];
  A[1] = A[H--];
  maxHeapify(1);
  return max;
}