Pat(Advanced Level)Practice--1051(Pop Sequence)

Pat1051代码

题目描述：

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

AC代码：

#include<cstdio>
#include<stack>

using namespace std;

int main(int argc,char *argv[])
{
	int m,n,k;
	scanf("%d%d%d",&m,&n,&k);
	while(k--)
	{
		int flag=1,cur=1;
		int i,temp;
		stack<int> s;
		for(i=1;i<=n;i++)
		{
			scanf("%d",&temp);
			if(flag)
			{
				while(s.empty()||s.top()!=temp)
				{
					s.push(cur);
					if(s.size()>m)
					{
						flag=0;
						break;
					}
					cur++;
				}
				if(flag&&s.top()==temp)
					s.pop();
			}
		}
		printf("%s\n",flag?"YES":"NO");
	}

	return 0;
}

如果栈顶元素与输入元素不相等，则一直压入元素，如果超出stack的大小则停止。

如果相等则弹出栈顶的元素。

如此进行下去....