PAT甲级1051

1051. Pop Sequence (25)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

#include<iostream>#include<stack>#include<vector>using namespace std;int main(){int m, n, k;cin >> m >> n >> k;stack<int> s; int t; vector<int> ts;for (int i = 0; i < k; i++){int num = 1; int count = 0;ts.clear();for (int j = 0; j < n; j++){cin >> t;ts.push_back(t);//注意读取输入流时在这个地方读完，不然后面中断时会导致有些数据没读完导致错误}while (1){count++;if (count > n)break;t = ts[count - 1];while (num <=t){s.push(num++);}if (s.size() > m){cout << "NO" << endl;break;}else{if (t != s.top()){cout << "NO" << endl;break;}elses.pop();}}if (s.empty()){cout << "YES" << endl;}while (!s.empty()){s.pop();}}return 0;}