The horse in the chessboard

There is a horse in the chessboard. Give its position. It has a destination. How many steps does it need to go to its destination.

input

• input: the first line has two integers: n,m, (n < = 1000, m < = 1000) indicating that the chessboard has n lines and m columns. The first row and the first column are (1,1)
• the second line: x1, y1, indicating the position of the horse.
• the third line, x2, y2, indicates its destination.
Ensure that the start and end positions are in the chessboard.
If the horse cannot reach the destination, output - 1

output

The minimum number of steps required for a horse to reach its destination

sample input

3 5
2 5
1 1

sample output

3

The code is as follows

#include<cstdio>
int quex[1000005],quey[1000005],ques[1000005];
int dr[8][2]={{-2,1},{-1,2},{1,2},{2,1},{2,-1},{1,-2},{-1,-2},{-2,-1}};
//dr is the direction array, quex, quey, ques is the queue array (can use structure), ques is the number of extended layers
bool book[1005][1005],flag;
//book marks the passing points, flag is used to determine whether to find, and exit the loop
int main()
{
    int n,m,x1,y1,x2,y2,xx,yy,head,tail;
    head=tail=1;
//The starting value of team head and team tail is 1
    scanf("%d%d",&n,&m);
    scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
    quex[tail]=x1;
    quey[tail]=y1;
    ques[tail]=0;
    tail++;
    //Enter the starting point coordinate into the team, and the tail points to the last element of the team
    book[x1][y1]=1;
    //Mark starting point
    while(head<tail)
    //Exit loop when team head and tail point to the same position
    {
        for(int i=0;i<8;i++)
        //Enumerate 8 directions
        {
            xx=quex[head]+dr[i][0];
            yy=quey[head]+dr[i][1];
            if(xx>0&&yy>0&&xx<=n&&yy<=m&&book[xx][yy]==0)
            {
                book[xx][yy]=1；
                //Mark the passing point
                quex[tail]=xx;
                quey[tail]=yy;
                ques[tail]=ques[head]+1;
                tail++;
                //Put the current coordinates in the team, and the tail points to the next element at the end of the team,
                //The number of layers is the number of layers of the node to be expanded + 1
            }
            if(xx==x2&&yy==y2){flag=1;break;}
            //When the horse reaches the end, exit the cycle
        }
        if(flag==1)break;
        //If found, exit the loop
        head++;
        //End of extension, change to the next extension
    }
    if(flag==0)printf("-1");
    else printf("%d",ques[tail-1])；
    //When flag=0, it is not found, because tail always points to the last element at the end of the team, so subtract 1
    return 0;
}