1.这道题是判断出栈队列是否合理。
2.采用了用栈来模拟情况。
3.当目前栈为空，或者栈不为空并且栈顶不等于目标值，并且队列中还有数值可以压入，栈的size小于最大容量，那么就一直循环执行压入操作，把123456789。。。队列中的值依次压入栈，直到跳出循环。
4.跳出循环后，判断栈顶是否等于目标值，不等于的话，这个sequence就是不合理的，如果等于，则把栈顶的值pop掉，再执行步骤3。
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
AC代码：
//#include<string>
//#include<stack>
//#include<unordered_set>
//#include <sstream>
//#include "func.h"
//#include <list>
#include <iomanip>
#include<unordered_map>
#include<set>
#include<queue>
#include<map>
#include<vector>
#include <algorithm>
#include<stdio.h>
#include<iostream>
#include<string>
#include<memory.h>
#include<limits.h>
#include<stack>
using namespace std;
int main(void)
{
int maxCap, maxSequence, querySum;
cin >> maxCap >> maxSequence >> querySum;
for (int k = 0; k < querySum; k++)
{
int idx = 1;
stack<int> sta;
int maxNum = maxCap;//栈里面能够存在的最大的值
vector<int> sequence(maxSequence);
for (int i = 0; i < maxSequence; i++)
scanf("%d", &sequence[i]);
bool result = true;
for (int i = 0; i < maxSequence; i++)
{
if (sequence[i]>maxNum)
{//不合理
result = false;
break;
}
while ((sta.empty() || (!sta.empty() && sta.top() != sequence[i])) && idx <= maxSequence && sta.size() <= maxCap)
{// 如果栈为空，或者栈不为空但是头部不等于目标值，并且还有值可以压入，队列size小于最大容量
sta.push(idx++);
}
if (sta.top() != sequence[i])
{//如果经过上面的压入操作后，仍不满足要求，则不合理
result = false;
break;
}
sta.pop();
maxNum++;//弹出了一个，可以压入更大的一个值
}
if (result)
cout << "YES" << endl;
else
cout << "NO" << endl;
}
return 0;
}