mooc浙大数据结构PTA习题之Pop Sequence

02-线性结构4 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
YESNo

参考代码（未评分）：（以下是我严重不满意的代码，运用了多重选择和循环，差点没看懂自己写的东西，有待改善。）

#include<iostream>
using namespace std;
const int stacksize =1024;
struct snode
{
	int data[stacksize];
	int top=-1;
};
void push(snode *p, int x)
{
	if (p->top ==stacksize - 1)
		cout << "栈已满" << endl;
	else
	{
		
		p->data[++(p->top)] = x;
	}
}
int pop(snode *p)
{
	if (p->top == -1)
		return 0;
	else
	{
		p->top--;

		return p->data[p->top + 1];
	}
}

void main()
{
	int M, N, K;
	int index, Y;
	int q,x;
	snode stack;
	snode *pp = &stack;

	cin >> M >> N >> K;
	int *a = new int[N*K];
	int *b = new int[2 * N+1];
	for (int m = 0; m <= N ; m++)
	{
		b[m] = m;
	}
	
	for (int i = 0; i < K; i++)
	{
		for (int j = 0; j < N; j++)
		{
			int k;
			cin >> k;
			a[j+i*N] = k;
		}
	}
	
		
	
		for (int i = 0; i < K; i++)
		{
			bool flag = 1;
			index = 0;
			Y = 1;
			int count = 0;
			int j = 0;
			for (int m = N + 1; m <= 2 * N; m++)
			{
				b[m] = 0;
			}

			for (int c = 0; c < N; c++)//结束内循环就要得出判断
			{
				
				while(index < Y)//当index<Y，将index+1`Y的值压入栈，注意栈的空间，抛出栈顶元素，和Y是否相等，若相等，更新index和Y的值，继续执行操作
				{
					
					for (q=index + 1;  q<=a[j+i*N]; q++)//入栈模拟
					{
						
						if (count>= M)//应结束入栈操作
						{
							break;
						}
						else
						{
							if (b[q + N] == 0)
							{
								push(pp, q);
								b[q + N] = 1;
								count++;
							}
							else
								continue;
						}
					}

						
					x = pop(pp);
					count--;

					if (x != a[j + i*N])
				    	        {
									flag = 0;
									break;
								}
					else//如果栈顶和当前值相等，则将当前值设为index,下一个数设为Y。需要比较index和Y的大小，来确定需要执行的操作。
								{
									index = a[j + i*N];
									if (j + 1 < N)
									{
										j++;
										Y = a[j + i*N ];
									}
									else
										break;

									continue;

								}
							}
				while(index>Y)
				{
					x = pop(pp);
					count--;
					if (x != Y)
					{
						flag = 0;
						break;
					}
					else
					{
						index = Y;
						if (j + 1 < N)
						{
							j++;
							Y = a[j + i*N ];
						}
						else
							break;

						 continue;



						}
						
					
				}

			}

			if (flag == 1)
				cout << "Yes" << endl;
			else
				cout << "No" << endl;

		}
	


	delete[]a;
	system("pause");
}