PAT 甲级 1051 Pop Sequence

1051 Pop Sequence （25 point(s)）

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

经验总结：

这一题，就是简单的模拟一下栈的出栈与入栈，能完成栈的模拟，且最终栈内为空的序列则为合法序列，其他的则为非法序列~

AC代码

#include <cstdio>
#include <cstring>
#include <iostream>
#include <stack>
using namespace std;
const int maxn=1010;
int num[maxn];
int main()
{
	int n,m,k;
	scanf("%d %d %d",&m,&n,&k);
	stack<int> s;
	for(int i=0;i<k;++i)
	{
		for(int j=0;j<n;++j)
			scanf("%d",&num[j]);
		int pos=0,flag=0;
		while(s.size())
			s.pop();
		for(int j=1;j<=n;++j)
		{
			s.push(j);
			if(s.size()>m)
			{
				flag=1;
				break;
			}
			while(s.size()&&s.top()==num[pos])
			{
				++pos;
				s.pop();
			}
		}
		if(flag==0&&s.size()==0)
			printf("YES\n");
		else
			printf("NO\n");
	}
	return 0;
}