解法一：模拟

根据栈顶和下一个待出栈元素的大小关系，决定进行出栈还是入栈操作，直到满足所给出栈顺序或者中间某一步无法实现。

#include <bits/stdc++.h>
using namespace std;
int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int M, N, K;
    cin >> M >> N >> K;
    deque<int> stack;
    while (K-- > 0) {
        stack.clear();
        int index = 1;
        int num;
        string flag = "YES\n";
        for (int i = 0; i < N; ++i) {
            cin >> num;
            // 保证非空
            if (stack.size() < M && stack.empty()) {
                stack.emplace_back(index++);
            }
            if (stack.back() == num) {
                stack.pop_back();
            } else if (stack.back() > num) {
                flag = "NO\n";
            } else {
                while (stack.size() < M && stack.back() < num) {
                    stack.emplace_back(index++);
                }
                if (stack.back() == num) {
                    stack.pop_back();
                } else {
                    flag = "NO\n";
                }
            }
        }
        cout << flag;
    }
}