1051 Pop Sequence (25 分)

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

分析：
该题的考点为栈的模拟。判断一个序列是否为正确的出栈序列，即模拟栈的入栈出栈过程。
用一个队列和一个栈来进行运算。将待检验的序列放在队列中，然后按照1-N的顺序入栈。每入栈一次就判断栈顶的元素和队列头部的元素是否相等。如果相等就一直将队列头部和栈顶的元素去掉，直到不相等。然后接着入栈，如此循环。最后判断栈是否为空：若栈为空，则序列合法，栈不为空，序列不合法。

#include <iostream>
#include <queue>
#include <stack>

using namespace std;

int main()
{
    
    int M, N, K, num;

    scanf_s("%d %d %d", &M, &N, &K);
    /*while (K--)
    {

    }*/
    for (int i = 0; i < K; i++)
    {
        stack<int> s;
        queue<int> q;
        for (int j = 0; j < N; j++) 
        {
            scanf_s("%d", &num);
            q.push(num);
        }
        for (int n = 1; n <= N; n++)
        {
            s.push(n);
            if (s.size() > M)
            {
                break;
            }
            while (!s.empty() && s.top() == q.front())
            {
                s.pop();
                q.pop();
            }
        }
        if (s.empty())
        {
            printf("YES\n");
        }
        else
        {
            printf("NO\n");
        }
    }
    system("pause");
    return 0;
}