1051. Pop Sequence (25)【栈】——PAT (Advanced Level) Practise_通过入栈和出栈求数列b-优快云博客

本文介绍了一道关于栈数据结构的编程题目，要求通过给定的容量和输入序列判断是否为有效的弹出序列。提供了完整的解题思路及AC代码。

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题目信息

1051. Pop Sequence (25)

时间限制100 ms
内存限制65536 kB
代码长度限制16000 B

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

解题思路

用stack模拟

AC代码

#include <cstdio>
#include <stack>
using namespace std;

int main()
{
    int m, n, k, t;
    scanf("%d%d%d", &m, &n, &k);
    while (k--){
        stack<int> st;
        int used = 0;
        bool flag = true;
        for (int i = 0; i < n; ++i){
            scanf("%d", &t);
            if (flag){
                while (st.empty() || st.top() != t){
                    st.push(++used);
                    if (st.size() > m){
                        flag = false;
                        break;
                    }
                }
                if (!st.empty() && st.top() == t){
                    st.pop();
                }
            }
        }
        printf("%s\n", flag ? "YES":"NO");
    }
    return 0;
}