1051. Pop Sequence

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最新推荐文章于 2022-11-05 17:32:25 发布

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分类专栏：数据结构和基本算法浙大PAT甲级真题及训练题文章标签： PAT 甲级浙大

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浙大PAT甲级真题及训练题

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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
#include <stdio.h>
#define MAXSIZE 1001
int top;                    //指针，始终指向栈顶元素，若栈为空，则为-1 
int main()
{
	int S[MAXSIZE];   //定义数组用作栈 
	int Max,n,k,i,j,m,num,num_max,flag;
	scanf("%d%d%d",&Max,&n,&k);   //依次读取栈的长度，序列中元素的个数，序列的个数 
	for (i=0;i<k;i++)       
	{
		top=-1;              //指针，始终指向栈顶元素，若栈为空，则为-1 
		flag=0;              //开关，初始为0 
		num_max=0;           //一个序列中已经读取到的最大元素，初始化为0 
		for (j=0;j<n;j++)
		{
			scanf("%d",&num); //读取一个元素 
			if (num>num_max)  //如果这个元素比已读取到最大元素还要大
			{
				for (m=num_max+1;m<=num;m++) //最大元素和这个元素之间的元素入栈 
				{
					S[++top]=m;
					if (top+1>Max) //如果入栈过程中超过栈的长度，则不存在这样的序列 
					{
						printf("NO\n");
						flag=1;
						break;
					}
				}
				num_max=num;      //重新将这个元素定义为最大元素 
				if (flag==1) break; 
				top--;
			}else      //如果这个元素比已读取到最大元素小，则将其与栈顶元素相比，看其是不是栈顶元素 
			{
				if (S[top--]!=num)    
				{
					printf("NO\n");
					flag=1;
					break;
				}
			}
		}
		if (flag==0) printf("YES\n");   //如果一个序列的所有元素都符合规则，则输出YES 
		while (j<n-1)     //如果循坏非正常结束，说明此序列不正确，把本序列剩下元素读取完，为下一个序列做准备 
		{
			scanf("%d",&num);
			j++;
		}   
	}
	return 0;
}

queue.pop

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03-24