PAT A 1051 Pop Sequence (25 分) 栈

最新推荐文章于 2024-04-25 10:28:26 发布

原创最新推荐文章于 2024-04-25 10:28:26 发布 · 234 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#PAT

PAT 专栏收录该内容

165 篇文章

订阅专栏

博客介绍了PAT A 1051题目的解题思路，即如何通过栈来模拟元素入栈和出栈操作，判断给定的序列是否可能是合法的出栈序列。文章提供了具体的解题方法，包括使用数组记录元素入栈状态，依次处理序列中的每个元素，检查栈容量和栈顶元素，以及相应的代码实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

1051 Pop Sequence (25 分)
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

题目大意：将元素1-N按1~N的顺序执行入栈和出栈操作，给出K个序列，判断它们是否是可能的pop序列。同时，栈的容量最大为M。

我的思路如下：记录每一组序列，然后使用stack来模拟这一过程。依次读取序列中的每个数，记为X，是应该被弹出的栈顶元素。为了避免元素重复入栈，使用数组vector inStack记录元素是否入过栈。每读入一个元素X，就将 1~X中所有未入栈的元素入栈，如果栈的大小大于M，则return false。之后判断栈顶元素是否是X，如果不是，则return false。如果正常执行了所有元素的读入，则return true。

代码如下：

#include<vector>
#include<iostream>
#include<cstdio>
#include<set>
#include<cmath>
#include<algorithm>
#include<climits>
#include<stack>
#pragma warning(disable:4996)

using namespace std;
int M, N, K;

bool judge(vector<int>& v)
{
	stack<int> st;
	vector<bool> instack(N+1, false);
	for (int i = 0; i < N; i++)
	{
		for (int j = 1; j <= v[i]; j++)
		{
			if (!instack[j])
			{
				st.push(j);
				instack[j] = true;
			}
		}
		if (st.size() > M) return false;
		int top = st.top();
		if (top != v[i]) return false;
		st.pop();
	}
	return true;
}

int main()
{
	scanf("%d%d%d", &M, &N, &K);
	for (int q = 0; q < K; q++)
	{
		vector<int> v(N);
		for (int i = 0; i < N; i++)
		{
			scanf("%d", &v[i]);
		}
		if (judge(v)) printf("YES\n");
		else printf("NO\n");
	}
	system("pause");
	return 0;
}