利用优先队列实现的小根堆排序；

#include <bits/stdc++.h>
#define int long long
#define pow_of_two(n) (n & -n == n) 
using namespace std;
#define lowbit(x) (x) & (-x)
// const int N = 1e6 + 10;

inline void solve() 
{   
    priority_queue<int , vector<int> , greater<int> > pq;
    int n , m;
    cin >> n >> m;
    for (int i = 1 , temp ; i <= n ; i ++ ) cin >> temp , pq.push(temp);

    while (m -- ) cout << pq.top() << " " , pq.pop();
    cout << endl;
    return ;
}


signed main()
{
    ios::sync_with_stdio(false) , cin.tie(0) , cout.tie(0);
    solve();
    return 0;
}

参考答案：

#include<iostream>

using namespace std;

const int N = 200010;
int a[N];
int n, m;

void down(int u)
{
    int t = u;
    if (2 * u <= n && a[2 * u] <= a[t]) t = 2 * u;
    if (2 * u + 1 <= n && a[2 * u + 1] <= a[t]) t = 2 * u + 1;

    if (t != u)
    {
        swap(a[u], a[t]);
        down(t);
    }
}
int main()
{
    cin >> n >> m;

    for (int i = 1; i <= n; ++i) cin >> a[i];

    for (int i = n / 2; i >= 1; --i) down(i);

    while (m--)
    {
        cout << a[1] << ' ';
        a[1] = a[n--];
        down(1);
    }

    return 0;
}