Pop Sequence

Pop Sequence

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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.

输入格式

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 T lines follow, each contains a pop sequence of N numbers. All the numbers in a line are separated by a space.

输出格式

For each pop sequence, print in one line “YES” if it is indeed a possible pop sequence of the stack, or “NO” if not.

输入示例

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

输出示例

YES
NO
NO
YES
NO

题目分析

题目大意是给定一个入栈序列1，2，3，…，N，限定栈的大小为***M***，要求判断序列是否有可能从该栈输出。

例如题目所给的样例：5 6 4 3 7 2 1：

1. 我们根据第一个出栈元素为5可以判断在5出栈前，1 2 3 4都应已入栈，则下一个应入栈元素是6。
2. 根据第二个元素为6可知，6出栈前栈里的元素为1 2 3 4 6；下一个进栈元素是7。
3. 根据第三个元素是4，第四个元素是3可知，7进栈前栈里的元素是1 2 ；7出栈后2，1分别出栈。整个过程符合1，2，3，…，N的入栈顺序，故该序列合法。

因此，若出现【栈顶元素大于待出栈的元素】或者【栈顶元素小于待出栈的元素】但【栈已满】的情况，则说明这个待出栈的元素是不可能出现的，据此可以判断该出栈序列不合法。

代码实现

链表栈

#include<iostream>
#include<stdlib.h>
using namespace std;
typedef struct node
{
	int *base;
	int top;
	int size;
}*stack;
int m, n, t;
stack CreateStack(int size)
{
	stack s = (stack)malloc(sizeof(struct node));
	s->base = (int*)malloc(size * sizeof(int));
	if (!s->base)
	{
		cout << "ERROR" << endl;
		return 0;
	}
	s->top = 0;
	s->size = size;
	return s;
}
void push(stack s,int e)
{
	if (s->top == s->size)//栈已满
	{
		cout << "FULL" << endl;
		return ;
	}
	s->base[++(s->top)] = e;
	return ;
}
void pop(stack s)
{	
	if (s->top == 0)//栈为空
	{
		cout << "EMPTY" << endl;
		return ;
	}
	s->top--;
}
int Check(int *array, stack s)
{
	int now = 1;
	int flag = 1;
	s->top = 0;//注意这里要重新设置栈顶位置
	for (int i = 0; i < n && flag; i++)
	{
 		while ((s->base[s->top] != array[i] || s->top == 0)&&flag)//【栈为空】或【栈顶元素与待比较的出栈元素不等】
		{
			push(s, now);
			if (s->top==s->size && s->base[s->top] != array[i]) //若【栈已满且栈顶元素仍然小于待比较的出栈元素】
			{
				flag = 0;
				break;
			}
			now++;
		}
		if ( flag && s->base[s->top] == array[i] && s->top >= 1)//若【栈非空且栈顶元素等于待比较的出栈元素】
			pop(s);
	}
	if (flag)
		cout << "YES" << endl;
	else
		cout << "NO" << endl;
	return 1;
}
int main()
{
	cin >> m >> n >> t;
	int *array = (int*)malloc(n * sizeof(int));
	stack s = CreateStack(m);
	while (t--)
	{
		for (int i = 0; i < n; i++)
			cin >> array[i];
		Check(array, s);
	}
	free(s->base);
	free(array);
	return 0;
}

数组栈

#include<iostream>
using namespace std;
int main()
{
	int m, n, t;
	cin >> m >> n >> t;
	int stack[1005];//栈
	while (t--)//遍历每一个序列
	{
		int top = 0;//栈顶元素
		int num;//待比较的出栈元素
		int flag = 1;//判断标志
		int now = 1;
		for (int i = 0; i < n; i++)//遍历每一个数
		{
			cin >> num;
			while ((top == 0 || num != stack[top]) && flag)//栈为空或者待出栈元素与栈顶元素不相等
			{
				stack[++top] = now;
				if (top > m)
				{
					flag = 0;
					break;
				}
				now++;
			}
			
			if (num == stack[top] && flag && top >= 1)//栈不为空且......
			{
				top--;
			}
		}
		if (flag)
			printf("YES\n");
		else 
			printf("NO\n");
	}
}