7-3 Pop Sequence (25分) Data Structures and Algorithms (English)

7-3 Pop Sequence (25分)

参考了博客 https://blog.csdn.net/whzyb1991/article/details/46663867
但是我看不懂(╥╯^╰╥)
题目链接 ：https://pintia.cn/problem-sets/16/problems/665
Given a stack which can keep M numbers at most. Push N numbers in the order of 1, 2, 3, …, N and pop randomly. You are supposed to tell if a given sequence of numbers is a possible pop sequence of the stack. For example, if M is 5 and N is 7, we can obtain 1, 2, 3, 4, 5, 6, 7 from the stack, but not 3, 2, 1, 7, 5, 6, 4.

Input Specification:
Each input file contains one test case. For each case, the first line contains 3 numbers (all no more than 1000): M (the maximum capacity of the stack), N (the length of push sequence), and K (the number of pop sequences to be checked). Then K lines follow, each contains a pop sequence of N numbers. All the numbers in a line are separated by a space.

Output Specification:
For each pop sequence, print in one line “YES” if it is indeed a possible pop sequence of the stack, or “NO” if not.

Sample Input:
5 7 5
1 2 3 4 5 6 7
3 2 1 7 5 6 4
7 6 5 4 3 2 1
5 6 4 3 7 2 1
1 7 6 5 4 3 2

Sample Output:
YES
NO
NO
YES
NO

作者: 陈越
单位: 浙江大学
时间限制: 400 ms
内存限制: 64 MB

#include
using namespace std;
typedef  long long ll;

int  main()
{
    int n,m,k,i;
    cin>>m>>n>>k;
    while(k--) {
        stack<int>s;
        vector<int>v;
        for (i = 0; i < n; i++) {
            int x;
            cin >> x;
            v.push_back(x);
        }
        s.push(0);//垫底的 这是看了另外一篇博客这样写的
        //这样就不用判断栈是不是空了 感觉挺好的
        // 一开始先放一个没什么用处的数据
        int idx = 0, fl = 0;
        int num = 1;
        for (idx = 0; idx < n; idx++) {//vector 的下标
            // 从左往右找我们要的数字  比如说 5 6 4 3 7 2 1  第一个要找的就是 5
            while (s.top() != v[idx]) {//如果栈顶不是我们要找的数字
                //就一直往栈里面放数字 题目要求是1 2 3 4 5  这样子放到
                // 就用num一直加一就行了
                s.push(num++);
                if (s.size() > m+1) {// 如果放进去这个数字之后，容量超出了 这是不行的
                    // 因为一开始放了个没用的数字进去 我们假定容量应该比题目给的 多一
                    // 也就是 m+1
                    fl = 1;
                    break;
                }
            }
            if (fl)break; // 前面发现他是超出容量的 直接退出了
              s.pop();// 找到了就把他弹出来

        }
        if (fl)printf("NO\n");
        else printf("YES\n");
    }

}
```cpp