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 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：m（栈的最大容量）   n（待压入序列的长度）   k（待压入序列的条数）

　　　　   往容量为m的栈里依次压入1~n的递增序列，输入k行序列，检验是否符合出栈规则，若符合输出“YES”，否则输出“NO”。

 

 1 #include 
 2 #include 
 3 using namespace std;
 4 int main() {
 5     int m,n,k;
 6     cin >> m >> n >> k;
 7     while(k--){
 8         int arr[1000];
 9         stack<int> s;
10         for(int i=0;i)
11             cin >> arr[i];
12         int current=0;
13         for(int i=0;i){
14             s.push(i+1);
15             if(s.size()>m) break;
16             while(!s.empty() && s.top()==arr[current]){
17                 s.pop();
18                 current++;
19             }
20         }
21         s.size()==0? cout << "YES" << endl : cout << "NO" << endl; 
22     }
23     system("pause");
24     return 0;
25 }

 

解法：每个序列的检验须经历以下三步：

　　   1、把输入的出栈序列存进数组，定义一个随时变量current作为数组指针；

　　　2、然后置一个栈并存进1~n的递增序列，每压入的一个数若栈顶元素等于数组指针指向的元素，则出栈并数组右移；

 　　  3、若栈为空说明该序列符合出栈序列。