PAT 1051 Pop Sequence (25分) 模拟入栈

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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;
}