题解:

相当于于对堆的基本应用
首先从2-n遍历，确定堆的类型
然后dfs后序遍历

Code:

#include <iostream>
#include <vector>
using namespace std;
int N, M;
vector<int> v;

// 后续遍历树
void dfs(int k)
{
    // 超过范围
    if (k > M)
        return;
    dfs(k * 2);
    dfs(k * 2 + 1);
    printf("%d%s", v[k], (k == 1 ? "\n" : " "));
}
int main()
{
    cin >> N >> M;
    while (N--)
    {
        v.clear();
        v.resize(M + 10);
        for (int i = 1; i <= M; i++)
            scanf("%d", &v[i]);
        int max_t = 1;
        int min_t = 1;

        // 先确定是什么堆
        for (int i = 2; i <= M; i++)
        {
            if (v[i] > v[i / 2])
                min_t = 0;
            if (v[i] < v[i / 2])
                max_t = 0;
        }
        // 同时满足不可能
        if (min_t == 0 && max_t == 0)
            cout << "Not Heap";
        else if (min_t == 0)
            cout << "Min Heap";
        else
            cout << "Max Heap";
        printf("\n");
        dfs(1);
    }
    return 0;
}