PAT甲级1051

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本文针对PAT甲级1051题进行详细解析，介绍如何判断给定序列是否为合法的栈出栈序列。通过设定栈容量及序列长度，利用C++实现逻辑判断，确保每个测试案例正确验证。

PAT甲级1051

原题：
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：给出栈容量，1~n个元素，待查的序列个数，判断给出的序列是否是合法的出栈序列

注意点：
1、每趟判断结束要清空栈！
2、边输入序列边判断，若是发现不合法不能直接break，可以立个flag标记，之后可以不做任何处理，但一定要继续把剩下数字读完！

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

int main(){
	int m,n,k;
	scanf("%d%d%d",&m,&n,&k);
	stack<int> s;
	for(int j=0;j<k;j++){
		int min=1;	//min记录下次进栈开始的数
		bool flag=true;	//记录是否是一个合法的出栈序列
		for(int i=0;i<n;i++){
			int t;
			scanf("%d",&t);
			if(flag){
				if(s.empty()||t>s.top()){	//若栈为空或者数字比栈顶元素大则进栈
					for(int l=min;l<=t;l++) s.push(l);
					min=t+1;	//下次进栈从t+1开始
					if(s.size()>m){	//若进栈的数量大于栈容量
						flag=false;
					}
				}
				if(t==s.top()) s.pop();
				else if(t<s.top()){	//若出栈的数字不是栈顶则出错
					flag=false;
				}
			}
		}
		if(flag) printf("YES\n");
		else printf("NO\n");
		while(!s.empty()) s.pop();	//清空栈
	}
	system("pause");
	return 0;
}