PAT1051 Pop Sequence (25分) 模拟入栈

题目

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

解析

题目：

有一个栈，大小为M，有N个数字，数字1~N只能按顺序依次入栈。给出K个出栈序列，判断该序列是否是可以实现的出栈序列。

思路：

入栈顺序是确定的（1-N依次入栈），我们只需要模拟这个过程。先把输入的序列接收进数组seq[]，设index= 1，表示当前比较的是序列哪个元素，然后按顺序1~n把数字进栈，每进入一个数字，判断有没有超过栈容量m，超过了就break。如果没超过，看看当前序列元素seq[index]是否与栈顶元素s.top()相等：while相等就一直弹出栈s.pop()，index++，相当于当前序列元素匹配成功，继续处理下一个；不相等就继续按顺序把数字压入栈。若能完美模拟，最终栈应该是空的，说明这个序列是可能的，输出YES

代码

#include <iostream> #include <stack> #include <vector> using namespace std; int main() { // 栈容量m，序列长度n，k个序列 int m, n, k; cin >> m >> n >> k; while (k -- > 0) { vector<int> seq(n + 1); stack<int> s; // 序列哪个位置元素 int index = 1; // 输入这个序列 for (int j = 1; j <= n; ++j) cin >> seq[j]; // 模拟入栈过程（只能从1到n入栈） for (int j = 1; j <= n; ++j) { s.push(j); // 栈中元素不能超过m个 if (s.size() > m) break; // 如果当前栈顶元素和当前序列元素相等 while (!s.empty() && s.top() == seq[index]) { // 出栈，相当于成功匹配这个元素 s.pop(); // 处理序列下一个元素 index++; } } // 如果最终栈空，说明这个序列可以是正确的出栈序列 if (s.empty()) cout << "YES" << endl; else cout << "NO" << endl; } return 0; }