A1051 Pop Sequence (25 point(s))

栈顺序已知，得不同出栈顺序，栈空间有限

1. 原文

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

2. 解析

3. AC代码

#include<cstdio>
#include<stack>
using namespace std;
int num[1010]={};
int main()
{
	int m,n,k;
	scanf("%d%d%d",&m,&n,&k);
	while(k--)
	{
		for (int i = 1; i <= n; ++i)
		{
			scanf("%d",&num[i]);
		}
		int current=1;
		stack<int> s;
		for (int i = 1; i <= n; ++i)
		{
			s.push(i);
			if (s.size()>m)
			{
				break;
			}
			while(!s.empty()&&s.top()==num[current]){
				s.pop();
				current++;
			}
		}
		if (current==n+1)
		{
			printf("YES\n");
		}else{
			printf("NO\n");
		}

		
	}
	return 0;
}