1051 Pop Sequence (25 分)

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

是有点恶心这题, 但是还是挺不错的一道题。

 1 #include <bits/stdc++.h>
 2 
 3 using namespace std;
 4 stack<int> st;
 5 int n,m,k;
 6 int an[1005];
 7 
 8 int main(){
 9     cin >> n >> m >> k;
10     for(int i = 0; i < k; i ++){
11         bool flag = true;
12         for(int j = 0; j < m; j ++){
13             cin >> an[j];
14         }
15         while(!st.empty()){
16             st.pop();
17         }
18         int p = 1;
19         for(int l = 0; l < m; l++){
20             if(p > m){
21                 flag = false;
22                 break;
23             }
24             // cout << p << endl;
25             if(an[l] != p){
26                 st.push(p++);
27                 // p++;
28                 if(st.size() > n){
29                     flag = false;
30                     break;
31                 }
32                 l--;
33             }else{
34                 st.push(p++);
35                 if(st.size() > n){
36                     flag = false;
37                     break;
38                 }
39                 while(!st.empty() && (l < m) && (an[l] == st.top())){
40                     // cout << st.top();
41                     st.pop();
42                     l++;
43                 }
44                 // if(st.empty())
45                 l--;
46             }
47         }
48         if(!st.empty()){
49             flag = false;
50         }
51         if(flag)
52             cout << "YES" << endl;
53         else
54             cout << "NO" << endl;
55     }
56     return 0;
57 }

 




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转载自www.cnblogs.com/zllwxm123/p/11185496.html
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