Pop Sequence (栈)

 Pop Sequence (栈)

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

一个数要出栈，那么比这个数小的数一定要先进栈 

 

设 tem 为进栈的数。如果 栈为空 或 出栈的数 不等于 栈顶数，那么 tem进栈 后 自增。如果 出栈的数 等于 栈顶数，则出栈。如果栈溢出，那就错误（可能是要出栈的数太大 或者是 要出栈的数 小于 此刻的栈顶元素 ）

#include <iostream>

#include <string>

#include<vector>

#include <stack>

using namespace std;

int main()

{

      int i,j,m,k,n,x;

      while(cin>>m)

      {

         cin>>n>>k;

         for(i=0;i<k;i++)

         {

          bool is=true;

              stack<int> tt;

              int tem=1;

              for(j=0;j<n;j++)

              {

              cin>>x;

                 while(tt.size()<=m && is)

                  {

                        if(tt.empty() || tt.top()!=x)

                              tt.push(tem++);

                        else if(tt.top()==x)

                        {

                            tt.pop();

                              break;

                        }

                  }

                  if(tt.size() > m)

                  {

                      is=false;

                  }

              }

               if(is) cout<<"YES"<<endl;

               else cout<<"NO"<<endl;

         }

      }

  return 0;

}