1051 Pop Sequence (25)

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

题目大意：给出一个栈的最大容量，依次向栈中压入1，2，3。。。；问给出的序列能否通过出栈实现
思路：模拟压栈，出栈；每次输入一个值 就判断当前值和栈顶的值是否，相同如果相同就弹出，否则一直压栈直到栈顶与输入值相同，当已经满栈，栈顶还是不和输入值相等就可以判断该序列不能通过出栈得到；
此外，当栈中只有一个值，且和输入值相同，弹出后栈为空，就需要压入一个值，否则下一次循环的比较会出错

#include<iostream>
#include<stack>
using namespace std;
int main(){
  int n, m, k, i, j;
  cin>>m>>n>>k;
  for(i=0; i<k; i++){
      bool flag=true;
      int cnt=2, temp;
      stack<int> s;
      s.push(1);
    for(j=0; j<n; j++){
          cin>>temp;
          if(s.top()==temp)s.pop();
          else{
              while(s.top()!=temp&&s.size()<m){s.push(cnt++);}
              if(s.top()==temp)s.pop();
              else flag=false;
          }
          if(s.empty())s.push(cnt++);
    }
    printf("%s
", flag?"YES":"NO");
  }
  return 0;
}