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>
#define SIZE 2000000
int h = 1, t = 1;

void maxHeapify(int *A, int i) {
    int l = 2 * i + h - 1 , r = l + 1, max;
    if (l < t && A[l] > A[i + h - 1]) max = l;
    else max = i + h - 1;
    if (r < t && A[r] > A[max]) max = r;

    if (max != i + h - 1) {
        int t = A[i + h - 1];
        A[i + h - 1] = A[max];
        A[max] = t;
        maxHeapify(A, max - h + 1);
    }
}

void buildMaxHeap(int *A) {
    int i;
    for (i = (t - h) / 2; i > 0; i--)
        maxHeapify(A, i);
}

int main(void) {
    int i, A[SIZE], k;
    char command[10];
    while (scanf("%s", command) != EOF) {
        if (command[2] == 's') {
            scanf("%d", &k);
            A[t++] = k;
            if (A[h] < k) buildMaxHeap(A);
        }
        else if (command[2] == 't') {
            //for (i = h; i < t; i++) printf("%d ", A[i]);
            printf("%d\n", A[h++]);
            buildMaxHeap(A);
        }
        else break;
    }

    return 0;
}