PAT甲级 1051. Pop Sequence (25)

题目：

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

思路：

这道题我是看：http://blog.youkuaiyun.com/u014646950/article/details/47430401这位作者的思路做的。一开始只想到了用数组和游标，没想到直接模拟出栈入栈的情况。

代码：

#include<iostream>
#include<stack>
#include<algorithm>
using namespace std;


int main()
{
	
	int M, N, K;
	cin >> M >> N >> K;

	int q[1001] = { 0 };
	int j,i = 0;
	
	for (; i < K; ++i)
	{
		for (j = 1; j <= N; ++j)
		{
			cin >> q[j];        //队列的第j位是q[j],输入顺序
		}
		//模拟入栈、出栈的过程
		stack<int> s;
		bool flag = 1;
		int index = 1;
		int dig = 1;
		while (flag && index <= N)
		{
			while (s.empty() || (q[index] != s.top()&& s.size()<M))
			{
				//把数字按顺序加到现在的数值，注意栈最大容量
				s.push(dig);
				dig++;
			}
			
			if (q[index] != s.top()) 
			{
				flag = 0;
			}
			else
			{
				s.pop();//踢出该数
				index++;
			}
		}
	
		if (index > N && flag)
		{
			cout << "YES" << endl;
		}
		else
			cout << "NO" << endl;
	}


	system("pause");
	return 0;
}