Pop Sequence

　　题目来源：PTA02-线性结构3 Pop Sequence   (25分)

　　Question: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:

　　Sample Output:

YES
NO
NO
YES
NO

　　分析：此题考察栈的操作，入栈的顺序是1,2,3......,N。出栈序列以5 6 4 3 7 2 1为例，要pop 5，就必须先push 1, push 2, push 3, push 4, push5, 此时栈顶元素为5，刚好匹配，才能进行pop操作。这里首先清空栈，设置一个将要入栈的顺序值temp，由1开始自增。当栈顶元素与读取的出栈序列值不匹配（还要考虑栈空的情况）时就进行入栈操作： Sta.push(temp++); ；当栈顶元素与读取的出栈序列值匹配，要进行出栈操作 Sta.pop(); 弹出栈顶元素，再读取下一个出栈序列值。如果栈中的元素个数超过了M，则说明出现了错误，这种出栈序列是不成立的。

　　源码：

#include<iostream>
#include<stack>        //调用C++ STL中的堆栈容器
using namespace std;

int main()
{
    int M, N, K;
    int obtain, pop;   // obtain为将要入栈的顺序值(1,2,..,N)，pop为当前读取的出栈序列值
    bool is_failed;    // 出栈序列成立与否的标志
    stack<int> sta;
    cin >> M >> N >> K;
    for (int i = ; i < K; i++)    　// 循环输入K组待判定的出栈序列
    {
        is_failed = false;
        obtain = ;
        for (int j = ; j < N; j++)  // 循环读取每个出栈序列值
        {
            cin >> pop;
            while (sta.size() <= M && !is_failed)    // 栈未满且未确认出栈序列不成立
            {
                if (sta.empty() || pop != sta.top()) // 栈为空或当读取的出栈序列值与栈顶元素不相等时，把顺序值temp压栈并递增
                {
                    sta.push(obtain++);
                }
                else      // 当前读取的出栈序列值与栈顶元素相等时出栈，跳出循环继续读取下一个出栈序列值
                {
                    sta.pop();
                    break;
                }
            }
            if (sta.size() > M)
            {
                is_failed = true;　　// 确认出栈序列不成立
            }
        }
        if (is_failed)    cout << "NO" << endl;
        else    cout << "YES" << endl;
        while (!sta.empty())    sta.pop();  // 清空栈，因为下一次匹配还要用
    }
    return ;
}