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
题目分析:一道模拟题 写了半天
1 #define _CRT_SECURE_NO_WARNINGS 2 #include <climits> 3 #include<iostream> 4 #include<vector> 5 #include<queue> 6 #include<map> 7 #include<set> 8 #include<stack> 9 #include<algorithm> 10 #include<string> 11 #include<cmath> 12 using namespace std; 13 stack<int> S1; 14 15 int main() 16 { 17 int M, N, K; 18 cin >> M >> N >> K; 19 vector<int> Array(N+1); 20 vector<int> Test(N+1); 21 for (int i = 1; i <= N; i++) 22 Array[i] = i; 23 for (int i = 0; i < K; i++) 24 { 25 while (S1.size()) 26 S1.pop(); 27 for (int j = 1; j <= N; j++) 28 cin >> Test[j]; 29 int k = 1, t = 1,flag=1; 30 while (t<=N) 31 { 32 if (S1.size()> M) 33 { 34 flag = 0; 35 break; 36 } 37 else if (S1.size()) 38 { 39 if (S1.top() == Test[t]) 40 { 41 S1.pop(); 42 t++; 43 } 44 else if(k<=N) 45 { 46 S1.push(Array[k]); 47 k++; 48 } 49 else 50 { 51 if (S1.top() == Test[t]) 52 { 53 S1.pop(); 54 t++; 55 } 56 else 57 { 58 flag = 0; 59 break; 60 } 61 } 62 } 63 else { 64 S1.push(Array[k]); 65 k++; 66 } 67 } 68 if (flag) 69 cout << "YES" << endl; 70 else 71 cout << "NO" << endl; 72 } 73 }