1051 Pop Sequence （25 分）

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 51 2 3 4 5 6 73 2 1 7 5 6 47 6 5 4 3 2 15 6 4 3 7 2 11 7 6 5 4 3 2

Sample Output:

YESNONOYESNO

 题目大意：给出一个栈最大容量，并且给出最大的N，要求是1....N是这样顺序入栈，输入K个检查的序列，需要判读是不是可能的弹出序列。

 //我看见其实以为自己会，其实是不会的，以前见过这种题的，但是应该当时也没理解吧。

 代码转自:https://www.liuchuo.net/archives/2232

 

#include 
     
      #include 
      
       #include 
       
        #include
        
         using namespace std;int main() {    int m, n, k;    scanf("%d %d %d", &m, &n, &k);    for(int i = 0; i < k; i++) {        bool flag = false;        stack
         
           s;        vector
          
            v(n + 1); for(int j = 1; j <= n; j++) scanf("%d", &v[j]);//读入要检验的序列。 int current = 1;//指向输入的序列。 for(int j = 1; j <= n; j++) { s.push(j); if(s.size() > m) break; while(!s.empty() && s.top() == v[current]) { s.pop(); current++; } } if(current == n + 1) flag = true; if(flag) printf("YES\n"); else printf("NO\n"); } return 0;}
          
         
        
       
      
     

 

//真的很厉害了，学习了。

1.使用一个current来指向当前的检验序列。

2.如果栈的大小已经大于了容量，那么就退出。

3.还有这个while循环是最关键的，只要top值等于当前current指向的，那么就弹出，并且指向下一个元素，非常厉害了。

//学习了，另一位大佬的代码思路也是相同的。