PAT 1051 Pop Sequence [模拟栈]

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

-----------------------------------这是题目和解题的分割线-----------------------------------

模拟进栈出栈的过程，我本来一边读取一边模拟，但出现了很多问题，就改成了先全部读入存取数组中再判断。

#include<cstdio>
#include<stack>

using namespace std;

int main()
{
	int m,n,k,i;
	scanf("%d%d%d",&m,&n,&k);
	while(k--)
	{
		int flag = 0,x,count = 0,a[1010];
		for(i=1;i<=n;i++)
			scanf("%d",&a[i]);
		stack<int> st;
		for(i=1;i<=n;i++)
		{
			//栈没满才可以入栈 
			while(st.size()<m)
			{	
				//直到栈顶元素和当前元素相等，不再入栈
				//这里得加是否为空的判断，不然碰到栈内没有元素还去判断是否相等的情况会有问题 
				if(!st.empty()&&st.top()==a[i]) break;
				st.push(++count); //按顺序入栈		
			}
			//如果是因为相等退出循环，出栈			
			if(a[i]==st.top()) st.pop();
			//其他原因，置flag = 1 
			else 
			{
				flag = 1;
				break;
			}	
		}
		if(flag) printf("NO\n");
		else printf("YES\n");
	}
	return 0; 
}