线性结构4 Pop Sequence

Posted 2020-10-18 A-Little-Nut

02-线性结构4 Pop Sequence（25 分）

Given a stack which can keep M numbers at most. Push N numbers in the order of 1, 2, 3, ..., Nand 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

 1 #include<iostream>
 2 #include<stack>
 3 #include<vector>
 4 #include<algorithm>
 5 using namespace std;
 6 int M,N,K;
 7 int check(vector<int> &vi){
 8     int m=0,n=0,cap=M+1;
 9     stack<int> sta;
10     sta.push(0); 
11     while(n<N){
12         while(sta.size()<cap&&vi[n]>sta.top())
13             sta.push(++m);
14         if(sta.top()==vi[n++]) sta.pop();
15             else return 0;
16         }    
17     return 1;
18 }
19 int main()
20 {
21     cin>>M>>N>>K;
22     vector<int> vi(N,0);
23     for(int j=0;j<K;j++){
24         for(int i=0;i<N;i++){
25         int n;
26         cin>>n;
27         vi[i]=n;}
28         if(check(vi)) 
29         cout<<"YES"<<endl;
30         else 
31         cout<<"NO"<<endl;
32         vi.clear();    
33     }
34     return 0;    
35  }

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