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

思路
栈的应用。这里直接用vector模拟栈，不得不说vector太强大了，既可以当栈又可以当队列用。

这道题需要理清楚弹出序列不满足的条件：
1.压进去数字后的栈容量大于限定的容量。
2.弹出序列需求的弹出元素不在栈顶。

简单过程：
1.用一个bool值isSuccess来标识弹出序列是否可行，默认为true。
2.保存第一次压入操作的最大元素cur，然后遍历弹出序列:
 1)如果弹出序列的当前元素temp比当前最大元素cur小，则它必须等于此时的栈顶元素top，否则要想得到和temp一样的值，就得不停弹出top，直到top == temp，但显然这样已经和弹出序列不匹配了，不可行，isSuccess置为false。
 2)如果temp比最大元素cur更大，则把cur + 1和temp之间的元素压入栈中，并检查此时栈容量是否超过要求，超过表明该序列也不可行,isSuccess置为false。

3.遍历完弹出序列后，根据标识isSuccess的值输出YES or No即可。

代码

#include<iostream>

#include<vector>

using namespace std;

int main()

{

     int N,M,K;

     while(cin >> M >> N >> K)

     {

         vector<int> stk;

         while(K--)

         {

           bool isSuccess = true;

           int curmax = ;

           stk.push_back();

           for(int i = ; i <= N;i++)

           {

               int temp;

               cin >> temp;

               if(temp > curmax)

               {

                   for(int j = curmax + ; j <= temp;j++)

                    stk.push_back(j);

                   if(stk.size() > M)

                   {

                       isSuccess = false;

                   }

                   curmax = temp;

               }

               else

               {

                   if(stk.back() != temp)

                 {

                   isSuccess = false;

                 }

               }

               stk.pop_back();

           }

            if(isSuccess)

                cout << "YES" << endl;

            else

                cout << "NO" << endl;

         }

     }

}