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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
#include<iostream> #include<stack> using namespace std; int main() { int M; //maximum capacity of the stack int N; //the length of push sequence int K; //the number of pop sequence to be checked cin >> M >> N >> K; int i, j; int input, temp; bool flag = true; stack<int> sta; for (i = 0; i < K; i++) { temp = 1; flag = true; for (j = 0; j < N; j++) { cin >> input; while (sta.size() <= M && flag) { if (sta.empty() || sta.top() != input) { sta.push(temp++); } else if (sta.top() == input) { sta.pop(); break; } } if (sta.size() > M) { flag = false; } } if (flag) cout << "YES" << endl; else cout << "NO" << endl; while (!sta.empty()) sta.pop(); } return 0; }
数据结构练习 02-线性结构3. Pop Sequence (25)
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原文地址：http://www.cnblogs.com/Zengineer/p/4538191.html
