Pop Sequence

Pop Sequence

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題目

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.

輸入格式

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.

輸出格式

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

輸入示例

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

輸出示例

YES
NO
NO
YES
NO

題目分析

題目大意是給定一個入棧序列1，2，3，…，N，限定棧的大小爲***M***，要求判斷序列是否有可能從該棧輸出。

例如題目所給的樣例：5 6 4 3 7 2 1：

1. 我們根據第一個出棧元素爲5可以判斷在5出棧前，1 2 3 4都應已入棧，則下一個應入棧元素是6。
2. 根據第二個元素爲6可知，6出棧前棧裏的元素爲1 2 3 4 6；下一個進棧元素是7。
3. 根據第三個元素是4，第四個元素是3可知，7進棧前棧裏的元素是1 2 ；7出棧後2，1分別出棧。整個過程符合1，2，3，…，N的入棧順序，故該序列合法。

因此，若出現【棧頂元素大於待出棧的元素】或者【棧頂元素小於待出棧的元素】但【棧已滿】的情況，則說明這個待出棧的元素是不可能出現的，據此可以判斷該出棧序列不合法。

代碼實現

鏈表棧

#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)//棧已滿
	{
		cout << "FULL" << endl;
		return ;
	}
	s->base[++(s->top)] = e;
	return ;
}
void pop(stack s)
{	
	if (s->top == 0)//棧爲空
	{
		cout << "EMPTY" << endl;
		return ;
	}
	s->top--;
}
int Check(int *array, stack s)
{
	int now = 1;
	int flag = 1;
	s->top = 0;//注意這裏要重新設置棧頂位置
	for (int i = 0; i < n && flag; i++)
	{
 		while ((s->base[s->top] != array[i] || s->top == 0)&&flag)//【棧爲空】或【棧頂元素與待比較的出棧元素不等】
		{
			push(s, now);
			if (s->top==s->size && s->base[s->top] != array[i]) //若【棧已滿且棧頂元素仍然小於待比較的出棧元素】
			{
				flag = 0;
				break;
			}
			now++;
		}
		if ( flag && s->base[s->top] == array[i] && s->top >= 1)//若【棧非空且棧頂元素等於待比較的出棧元素】
			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;
}

數組棧

#include<iostream>
using namespace std;
int main()
{
	int m, n, t;
	cin >> m >> n >> t;
	int stack[1005];//棧
	while (t--)//遍歷每一個序列
	{
		int top = 0;//棧頂元素
		int num;//待比較的出棧元素
		int flag = 1;//判斷標誌
		int now = 1;
		for (int i = 0; i < n; i++)//遍歷每一個數
		{
			cin >> num;
			while ((top == 0 || num != stack[top]) && flag)//棧爲空或者待出棧元素與棧頂元素不相等
			{
				stack[++top] = now;
				if (top > m)
				{
					flag = 0;
					break;
				}
				now++;
			}
			
			if (num == stack[top] && flag && top >= 1)//棧不爲空且......
			{
				top--;
			}
		}
		if (flag)
			printf("YES\n");
		else 
			printf("NO\n");
	}
}