1051. Pop Sequence (25)

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.

5 7 51 2 3 4 5 6 73 2 1 7 5 6 47 6 5 4 3 2 15 6 4 3 7 2 11 7 6 5 4 3 2Sample Output:
YESNONOYESNO
模拟进栈出栈，注意栈大小限制
#include<cstdio>#include<stack>using namespace std;const int maxn = 1010;int a[maxn];stack<int> st;int main(){	int n, m, k;	scanf("%d %d %d", &m, &n, &k);	while(k--){		for(int i = 1; i <= n; ++i){			scanf("%d", &a[i]);		}		while(!st.empty()){			st.pop();		}		int pop_seq = 1;   //入栈元素push_now， 出栈序列索引pop_seq		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() == a[pop_seq]){				st.pop();				++pop_seq;			}		}		if(st.empty() && flag ){			printf("YES/n");		}else{			printf("NO/n");		}	}	return 0;}