7-9 one way in, two ways out (25 points)

Consider a special queue which is a linear structure that allows insertions at one end, yet deletions at both ends. Your job is to check, for a given insertion sequence, if a deletion sequence is possible. For example, if we insert 1, 2, 3, 4, and 5 in order, then it is possible to obtain 1, 3, 2, 5, and 4 as an output, but impossible to obtain 5, 1, 3, 2, and 4.

Input Specification:

Each input file contains one test case. For each case, the first line gives 2 positive integers N and K (≤10), which are the number of insertions and the number of queries, respectively. Then N distinct numbers are given in the next line, as the insertion sequence. Finally K lines follow, each contains N inserted numbers as the deletion sequence to be checked.

All the numbers in a line are separated by spaces.

Output Specification:

For each deletion sequence, print in a line yes if it is indeed possible to be obtained, or no otherwise.

Sample Input:

5 4
10 2 3 4 5
10 3 2 5 4
5 10 3 2 4
2 3 10 4 5
3 5 10 4 2

Sample Output:

yes
no
yes
yes

Chinese translation ： 

Consider a special queue , It's a linear structure , Allow one end to insert , Both ends are deleted . Your job is to check whether a given insertion sequence can delete the sequence . for example , If we insert in order 1、2、3、4 and 5, You can get 1、3、2、5 and 4 As the output , But it is impossible to get 5、1、3、2 and 4.

Enter specifications ：

Each input file contains a test case . For each case , The first line gives 2 A positive integer N and K(≤10） , Insert number and query number respectively . Then in the next line N A different number , As an insert sequence . And finally K That's ok , Each row contains N Insert the number as the deletion sequence to be checked .

All numbers in a row are separated by spaces .

Output specifications ：

For each deletion sequence , If you can really get , Then print in the line “ yes ”, Otherwise print “ no ”.

Sample input ：

First of all, let's understand the meaning of the question , For example 1234 Is the default sequence , Then he gave a 4123 This sequence , So it's according to 4123 Delete the data in the order ,4 There must be before 123,3 There must be before 12........, Then insert only one end , First, 4, Then we insert from one end 1234, Then delete both ends , First, 4, Delete from the right , And then there was 1, Delete from the left , And then there was 2, Delete from the left , And then there was 3 Delete from the right . At the same time, if a number is deleted , Then we don't need to consider this number in the later sequence , For example 2 Deleted , that 3 As long as there is 1 that will do .

From the above example, we first judge a number , What numbers did he have to appear first , Now we need to construct a sequential function , That is, before a certain number appears , Those numbers should appear , meanwhile , We also need an array to mark whether a number has been deleted , It is also necessary to determine whether the numbers at both ends are corresponding （ That is, the number we want to delete ）, If not, then NO. According to this, we write the following code ：

#include <stdio.h>
int main(){
	int i,j;
	int k,n,front,flag,m,t,rear;
	scanf("%d%d",&n,&k);
	int a[n],c[n],b[n],book[n],g,h,l;//c[n] Is to store the current undeleted number ,a[n] It's a sequential array ,b[n] Is the order in which we want to delete numbers .
	for(i=0;i<n;i++){
		scanf("%d",&a[i]);
		book[i]=1;// All data has not been deleted 
		c[i]=-1;// Prevent random data from being duplicated with the data we want to delete 
	}
	for(i=0;i<k;i++){
		front=0;
		flag=0;
		l=0;
		rear=-1;// Adding a data is from 0 The subscript begins , So we let rear yes -1
		for(j=0;j<n;j++)book[j]=1;
		for(j=0;j<n;j++){
			h=0;
			scanf("%d",&b[j]);// Enter the data to be added 
			for(g=0;g<n;g++)if(a[g]==b[j])break;// Find the data in a Position in array , It's convenient to save the data in front of him first 
			if(l<=g){for(m=0,l=0;l<=g;l++){// If the data to be stored is larger than the previous data , We need to update the array .
				if(book[l]==0)continue;// If it has been deleted, it will not be executed 
				c[h++]=a[l];
				front=0;// Because the array is going to 0 Inserted above , So we let the head become 0.
			}
			rear=h-1;
		}
			if(c[front]==b[j]){// See whether the header is the data to be deleted 
				book[g]=0;
				front++;// Delete the head , Because there is no deletion in memory , So we need to logically let front++, To achieve logical deletion .
			}
			else if(c[rear]==b[j]){// See whether the tail is the data to be deleted 
				book[g]=0;
				rear--;
			}
			else {j++;// Neither is it no
			flag=1;
			break;
			}
		}
		for(t=j;t<n;t++)scanf("%d",&b[t]);// because break, As a result, some data is not scanf, So here is another loop to accept the data .
		if(flag)printf("no\n");
		else printf("yes\n");
	}
}

  The code I wrote just logically deleted the data , But the data is not deleted completely in the memory .