【C++】PAT(advanced level)1051. Pop Sequence (25)_given a stack which can keep m numbers at most. pu-优快云博客

本文介绍了一个算法问题：如何判断给定的序列是否为特定容量栈的有效弹出序列。通过使用C++实现，文章详细解释了如何在限定的时间和内存内解决此问题，并提供了具体的输入输出示例。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

1051. Pop Sequence (25)

时间限制

10 ms

内存限制

32000 kB

代码长度限制

16000 B

判题程序

Standard

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

1.stack的用法

啊啊

//#include<stdio.h>
#include<iostream>
#include<algorithm>
//#include<string.h>
#include<vector>
//#include<map>
#include<stack>
//#include<iomanip>
using namespace std;

int main(){
	freopen("in.txt","r",stdin);
	int M,N,n;
	int cc=1;
	bool ff=true;
	cin>>M>>N>>n;
	
	while(n--){
		stack <int> s;
		int m=N;
		cc=1;
		ff=true;
		while(m--){
		
			int num;
			scanf("%d",&num);
			while(cc<=num){
				if(s.size()==M){
					ff=false;
					break;
				}
				s.push(cc);
				cc++;
			}
			if(s.top()!=num||!ff){
				ff=false;
				continue;
			}
			s.pop();
		}
		if(ff){
			cout<<"YES"<<endl;
		}else{
			cout<<"NO"<<endl;
		}
	}

	fclose(stdin);
	system("pause");
	return 0;
}