菜鸡奋斗路02-线性结构4 Pop Sequence

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

作者: 陈越

单位: 浙江大学

时间限制: 400ms

内存限制: 64MB

代码长度限制: 16KB

---

个人分析：这道题考察堆栈的基本操作（入栈出栈），题目要求，在入栈增序（123...）的前提下，判断一系列出栈样例，若样例是可能获得的出栈样例则输出YES，否则输出NO。菜鸡第一个想法，就是自己先手动试试，怎样的出栈pop序列是不可能的，如图:

嗯，给它想了一个很霸气的名字——动态出入栈，也就是根据所需判断的样例进行入栈和出栈操作，从而判断样例是否符合条件。时间有点晚，菜鸡忙着写完博客去学习（其实是吃鸡），这就不多废话了，上代码！！

#include<stdio.h>
#include<stdlib.h>
#define Max 1000
//采用链表存储堆栈
typedef struct Stack *List;
struct Stack{
	int data;
	List next;
}; 

bool IsEmpty(List L)
{
	return(L->next==NULL);
}

bool Push(int size,List L1,int data)
{
	int cnt=0;
	List p=L1;
	while(p->next!=NULL)
	{
		cnt++;
		p=p->next;
	}
		
	if(cnt>=size)
	{
		return false;
	}
	else
	{
		List T;
		T=(List)malloc(sizeof(struct Stack));
		T->data=data;
		T->next=L1->next;
		L1->next=T;
		return true;
	}
}

void Pop(List L2)
{
	List curr;
	curr=L2->next;
	L2->next=curr->next;
	free(curr);
}

int main()
{
	//size堆栈大小，number需入栈元素个数，exam测试样例数目，sample测试数据
	int size,number,exam,sample;  
	scanf("%d %d %d\n",&size,&number,&exam);
	int FLAG[Max]={0};
	//FLAG[]记录每个测试样例的flag，即记录每个测试样例是否符合题目条件
	 for(int i=0;i<exam;i++)
	 {	
	 	List S;
		S=(List)malloc(sizeof(struct Stack));
		S->next=NULL;
	 	int flag=0;   //判断输入样例是否符合条件，不符合为1，符合为0 
	 	int t=1;
	 	for(int j=0;j<number;j++)
	 	{	                     //动态入栈，根据输入序列控制入栈出栈
	 		if(IsEmpty(S))       //同时判断输入序列是否符合pop（出栈）规则
	 		{
	 			Push(size,S,t);     
	 			t++;
			}	
		 	scanf("%d",&sample);
	 		List temp=S->next;
	 		while(temp->data<sample)
			{
				if(!Push(size,S,t))
				{
					flag=1;
					break;
				}
				else
				{
					temp=S->next;
				} 	
				
				t++;	
			}
			if(temp->data==sample)
			{
				Pop(S);
				temp=S->next;
			}
			else if(temp->data>sample)
			{
				flag=1;
			}						
		}
		if(!flag)
			FLAG[i]=0;
		else
			FLAG[i]=1;
		getchar();
	 }
	 for(int i=0;i<exam;i++)
	 {
	 	if(FLAG[i])
	 		printf("NO\n");
	 	else
	 		printf("YES\n");
	 }
	 return 0;
} 

测试结果：

OKOKOK，菜鸡要开始今天的happy time了！！吃鸡的吃鸡，王者荣耀的王者荣耀！！GOGOGO！