PAT甲级1051 Pop Sequence (25 分)

1051 Pop Sequence (25 分)

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

      题目大意：

         给出容量为m的栈，顺序入栈。现有k条长度为n的序列，试问是否可能是出栈序列。

 

思路：

         进行简单模拟，读入出栈序列，然后循环执行只要栈不为空且栈顶与出栈队列相同，则出栈。判断是否为出栈序列只要查看最后是否有把出栈队列读完即可。

 

参考代码：

#include<vector>
#include<stack>
#include<cstdio>
using namespace std;
int m, n, k;
vector<int> num;
int main(){
	scanf("%d%d%d", &m, &n, &k);
	num.resize(n);
	for(int i = 0; i < k; ++i){
		for(int j = 0; j < n; ++j)	scanf("%d", &num[j]);
		int idx = 0;
		stack<int> st;
		for(int j = 1; j <= n; ++j){
			st.push(j);
			if(st.size() > m)	break;
			while(!st.empty() && st.top() == num[idx])	st.pop(), idx++;
		}
		if(idx == n)	printf("YES\n");
		else	printf("NO\n");
	}
	return 0;
}