【PAT A1051】Pop Sequence

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.

#include <cstdio>
#include <stack>
using namespace std;

int main(){
	int M, N, K, seq[1010];
	scanf("%d%d%d", &M, &N, &K);
	while(K--){
		for(int i = 0; i < N; i++)
			scanf("%d", &seq[i]);
		stack<int> s;
		//i是下一个push的数，j是即将验证的数
		int i = 1, j = 0;
		while(j < N){
            //当要验证的数比要push的数大而且栈里尚有空间则入栈
			if(i <= seq[j] && s.size() < M){
				s.push(i);
				i++;
			}
			else{
                //检查栈顶是否是要验证的数，是则出栈否则说明该栈序列不可能
				if(s.top() == seq[j]){
					s.pop();
					j++;
				}
				else{
					printf("NO\n");
					break;
				}
			}
		}
        //j=N说明已经验证完序列中所有的数
		if(j == N)
			printf("YES\n");
	}
	
	return 0;
}