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 JAVA (judge verdict: Time Limit Exceeded)

import java.io.*;
import java.util.*;

public class Main {
	static int H=0;
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int A[]=new int[2000001];
		
		for(int i=0;i<2000001;i++)A[i]=-1;
		
		PrintWriter out=new PrintWriter(System.out);
		while(true) {
			String odr = in.next();
			if(odr.equals("end"))break;
			else if(odr.equals("insert")) {
				int k=Integer.parseInt(in.next());
				A[H+1]=k;
				H++;
				buildMaxHeap(A);
			}
			else if(odr.equals("extract")) {
				out.println(A[1]);
				delTop(A);
			}
		}
		out.flush();
	}
	
	static void maxHeapify(int A[],int index) {
		int l=index*2, r=index*2+1, L;
		
		if(l<=H && A[l]>A[index])L=l;
		else L=index;
		if(r<=H && A[r]>A[L])L=r;
		
		if(L!=index) {
			int t=A[index];
			A[index]=A[L];
			A[L]=t;
			maxHeapify(A,L);
		}
	}
	
	static void delTop(int A[]) {
		A[1]=-1;
		for(int i=1;i<H;i++) {
			int t=A[i+1];
			A[i+1]=A[i];
			A[i]=t;
		}
		H--;
		buildMaxHeap(A);
	}
	
	static void buildMaxHeap(int A[]) {
		for(int i=H/2;i>0;i--)maxHeapify(A,i);
	}

}