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 2000000

int n,A[MAX+1];
void insert(int);
void increase(int,int);
void max(int);
int extract();

int main(){
    int k;
    char CHAR[10];

    while(1){
        scanf("%s",CHAR);
        if(strcmp(CHAR,"end")==0) break;
        if(strcmp(CHAR,"insert")==0){
            scanf("%d",&k);
            insert(k);
        }
        else{
            printf("%d\n",extract());
        }
    }
    return 0;   
}

void insert(int k){
    n++;
    A[n]=-1;
    increase(n,k);
}
void increase(int n,int k){
    int temp;
     if(k<A[n]) return;
     A[n]=k;
     while(n>1 && A[n/2]<A[n]){
         temp=A[n];
         A[n]=A[n/2];
         A[n/2]=temp;
         n=n/2;
     }
}
void max(int i){
    int l,r,larg,temp;
    l=2*i;
    r=2*i+1;
    if(l<=n && A[l]>A[i]) larg=l;
    else larg=i;
    if(r<=n && A[r]>A[larg]) larg=r;

    if(larg!=i){
       temp=A[i];
       A[i]=A[larg];
       A[larg]=temp;
       max(larg);
    }
}
int extract(){
    int m;
    if(n<1) return 0;
    m=A[1];
    A[1]=A[n];
    n--;
    max(1);
    return m;
}