搜索--A Knight's Journey

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

代码：

#include
#include
#include
using namespace std;
int maxn=10010;
int v[35][35],v1[35][35];
char vis[905][2];
int f[8][2]={{-1,-2},{1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}};
int l=1,n,l1,l2,flag,t=1;
int Judge(int x1,int y1)
{
    if(x1<0||x1>=l1||y1<0||y1>=l2)return 0;
    if(v[x1][y1]>0)return 0;
    return 1;
}
void dfs(int x,int y)
{
    vis[l][0]=x+'1';
    vis[l][1]=y+'A';
    if(l==l1*l2){
        flag=1;
        for(int i=1;i<=l1*l2;i++){
        printf("%c%c",vis[i][1],vis[i][0]);
        }
        printf("\n");
        return;
    }
    for(int k=0;k<8&&(flag==0);k++){
        int x1,y1;
        x1=x+f[k][0];
        y1=y+f[k][1];
        if(Judge(x1,y1)){
            v[x1][y1]=1;
            l++;
            dfs(x1,y1);
            l--;
            v[x1][y1]=0;
        }
    }
    return;
}
int main(){
    scanf("%d",&n);
    while(n--){
    scanf("%d %d",&l1,&l2);
    printf("Scenario #%d:\n",t++);
            l=1;   flag=0;
            v[0][0]=1;
            dfs(0,0);
            memset(v,0,sizeof(v));
            memset(v1,0,sizeof(v1));
            memset(vis,0,sizeof(vis));
    if(!flag) printf("impossible\n");
    if(n) printf("\n");
    }
    return 0;
}

在搜索是直接从（0，0）开始就行，在某一点存在不能搜索的点，在任何地方都会有不能搜索到的点。
横纵坐标。。。。。。怎么分辨
字典序要求搜索时

 f[8][2]={{-1,-2},{1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}};

一定是向坐标最小的地方走