1051. Pop Sequence

快来打我* 2022-06-16 10:07 191阅读 0赞

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;
    }