PAT甲级——Pop Sequence

最新推荐文章于 2022-03-01 16:06:52 发布

Xiao2018428

最新推荐文章于 2022-03-01 16:06:52 发布

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本文介绍了一种算法，用于判断给定的出栈序列是否可能由一个最大容量为M的栈通过一系列的推入和随机弹出操作得到。通过模拟入栈和出栈过程，文章详细解释了如何检查出栈序列的有效性。

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

思路：模拟入栈出栈的过程，将1、2.....n依次入栈，在入栈的过程中如果入栈的元素恰好等于出栈序列当前等待出栈的元素，那么就让栈顶的元素出栈，同时把出栈序列当前等待出栈的元素位置后移1位。此时只要栈顶元素仍然等于出栈序列当前等待出栈的元素，则持续出栈。

样例1:

1入栈，1出栈，2入栈，2出栈，3入栈，3出栈，4入栈，4出栈，5入栈，5出栈，6入栈，6出栈，7入栈，7出栈。序列符合切没有超过栈的最大容量，故输出“YES”。

样例2:

1入栈，2入栈，3入栈，3出栈，2出栈，1出栈，4入栈，5入栈，6入栈，7入栈，7出栈，但是此时栈顶元素为6，所以接下来5出栈的命令不可能成功执行，输出“NO”。

代码：

#include <iostream>
#include<stack>
using namespace std;
const int maxn = 1010;
int arr[maxn];//保留出栈队列
stack<int>st;

int main()
{
	int m, n, T;
	scanf("%d%d%d", &m, &n, &T);
	while (T--)
	{
		while (!st.empty())
		{
			st.pop();
		}
		for (int i = 1; i <= n; i++)
		{
			scanf("%d", &arr[i]);
		}
		bool flag = true;
		int current = 1;
		for (int i = 1; i <= n; i++)
		{
			st.push(i);//把i压入栈
			if (st.size() > m)
			{
				flag = false;
				break;
			}
			while(st.empty() == false && st.top() == arr[current])
			{
				st.pop();
				current++;
			}
		}
		if (flag == true && st.empty() == true)
		{
			printf("YES\n");
		}else{
			printf("NO\n");
		}
	}
	return 0;
}