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: Time Limit Exceeded)


#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>


int parent(int);
int left(int);
int right(int);

void maxHeapify(int*,int,int);

void insert(int*,int);
int extractMax(int*,int);



int main(){
  int i,j,n;
  int max;
  char order[10];
  int num[100000];  

  i=1;

  while(1){
    scanf("%s",order);
    if(strcmp(order,"insert")==0){
      insert(num,i);
      i++;
    }
    if(strcmp(order,"extract")==0){
      max=extractMax(num,--i);
      printf("%d\n",max);
    }
    if(strcmp(order,"end")==0) break;
  }  
  return 0;

}

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

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

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

void insert(int *num,int i){
  int n,j;
  scanf("%d",&n);
  num[i]=n;
  for(j=i/2;j>=1;j--){
  maxHeapify(num,j,i);
  }
}
void maxHeapify(int *num,int i,int n){//並びかえ
  int r,l;
  int largest;
  int tmp;
  l=left(i);
  r=right(i);

  if(l<=n && num[l]>num[i]){
    largest=l;
  }
  else{ largest=i; }

  if(r<=n && num[r]>num[largest]){
    largest=r;
  }

  if(largest!=i){
    tmp=num[i];
    num[i]=num[largest];
    num[largest]=tmp;
    maxHeapify(num,largest,n);
  }
    
}
int extractMax(int *num,int i){
  int max;

  max=num[1];
  num[1]=num[i];
  --i;
  maxHeapify(num,1,i);
  return max;
}