dfs:Knight's Journey

Description
Background
The knight is getting bored of seeing the same black and white squares again and again and has decided to make a journey
around the world. Whenever a knight moves, it is two squares in one direction and one square perpendicular to this. The world of a knight is the chessboard he is living on. Our knight lives on a chessboard that has a smaller area than a regular 8 * 8 board, but it is still rectangular. Can you help this adventurous knight to make travel plans?

Problem
Find a path such that the knight visits every square once. The knight can start and end on any square of the board.
Input
The input begins with a positive integer n in the first line. The following lines contain n test cases. Each test case consists of a single line with two positive integers p and q, such that 1 <= p * q <= 26. This represents a p * q chessboard, where p describes how many different square numbers 1, … , p exist, q describes how many different square letters exist. These are the first q letters of the Latin alphabet: A, …
Output
The output for every scenario begins with a line containing “Scenario #i:”, where i is the number of the scenario starting at 1. Then print a single line containing the lexicographically first path that visits all squares of the chessboard with knight moves followed by an empty line. The path should be given on a single line by concatenating the names of the visited squares. Each square name consists of a capital letter followed by a number.
If no such path exist, you should output impossible on a single line.
Sample Input
3
1 1
2 3
4 3
Sample Output
Scenario #1:
A1

Scenario #2:
impossible

Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4

#include 
#include 
using namespace std;

int n,m,vis[300][300],p[1009][1009],flag;
int dir[8][2]= {{-1,-2},{1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}};

void dfs(int x,int y,int index)
{
    if(index==n*m)
    {
        flag=1;
        for(int i=0; i<index; i++)
            printf("%c%d",'A'+p[i][1],p[i][0]+1);
        printf("\n");
    }
    else
    {
        for(int i=0; i<8; i++)
        {
            int xx=x+dir[i][0];
            int yy=y+dir[i][1];
            if(xx>=0&&xx=0&&yy<m&&!vis[xx][yy])
            {
                vis[xx][yy]=1;
                p[index][0]=xx;
                p[index][1]=yy;
                dfs(xx,yy,index+1);
                vis[xx][yy]=0;
                if(flag)
                    break;
            }
            if(flag)
                break;
        }
    }

}

int main()
{
    int t;
    scanf("%d",&t);
    p[0][0]=0;
    p[0][1]=0;
    for(int k=1; k<=t; k++)
    {
        flag=0;
        memset(vis,0,sizeof(vis));
        scanf("%d%d",&n,&m);
        printf("Scenario #%d:\n",k);
          vis[0][0]=1;
        dfs(0,0,1);
        if(!flag)
            printf("impossible\n");
        if(kprintf("\n");
    }
    return 0;
}

Reflect:
1，由于需要字典序最小，所以方向数组的顺序很重要
2，在记录数组上不用担心覆盖问题，因为index的存在！只要将vis置零就可以；