A1051 Pop Sequence(25점)[C 언어]

A1051 Pop Sequence(25점)[C 언어]
원제 링크
실제 입고, 출고를 시뮬레이션하여 문제에서 제시한 창고 서열이 정확한지 판단합니다
제목 설명: 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.
입력 형식: 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.
출력 형식: For each pop sequence, print in one line "YES"if it is indeed a possible pop sequence of the stack, or "NO"if not.
샘플 입력:
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
출력 예제:
YES NO NO YES NO
구현 코드:

#include 
#include 
using namespace std;
const int maxn = 1010;
int arr[maxn]; //          
stack<int> st; // st，  int    

int main()
{
	int m, n, k;
	scanf("%d%d%d", &m, &n, &k); 
	while(k--){
		while(!st.empty()){ //    
			st.pop();
		}
		for(int i=1; i<=n; ++i){
			scanf("%d", &arr[i]);
		}
		int current = 1; //                   
		bool flag = true; //           
		for(int i=1; i<=n; ++i){
			st.push(i);
			if(st.size()>m){ //              ，             
				flag = false;
				break;
			}
			while(!st.empty() &&st.top()==arr[current]){ //             
				//                  
				st.pop();
				current++; //          
			}
		}
		if(st.empty() && flag){ //          
			printf("YES
");
		}else{
			printf("NO
");
		}
	}
	
	return 0;
}