poj 2488 A Knight's Journey

A Knight's Journey

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 25609		Accepted: 8735

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<stdio.h>
#include<string.h>
struct node
{
     int x;
     int y;
}pos[10000];

int dir[8][2]={{-1,-2},{1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}};
//貌似方向数组要这样写，题目要求按字典序输出
int n,m,vis[40][40],flag;
void  dfs(int x,int y,int num)
{
     int i;
     vis[x][y]=1;
     if(num==n*m)
     {
          flag=1;
          return ;
     }
     if(flag)
     return ;

     for(i=0;i<8;i++)
     {
                 if(flag)
               break;
          int xx=x+dir[i][0],yy=y+dir[i][1];
          if(xx>=0&&xx<n&&yy>=0&&yy<m&&!vis[xx][yy])
          {
               vis[xx][yy]=1;
               pos[num].x=xx;
               pos[num].y=yy;
               dfs(xx,yy,num+1);
               vis[xx][yy]=0;


          }
     }
     return ;
}
int main()
{
     int i,j,t,cas=1;
     //freopen("in.txt","r",stdin);
     //freopen("out.txt","w",stdout);
     scanf("%d",&t);
     while(t--)
     {
          flag=0;
          memset(vis,0,sizeof(vis));
           memset(pos,0,sizeof(pos));
          scanf("%d %d",&n,&m);
          pos[0].x=0;
          pos[0].y=0;
          dfs(0,0,1);//貌似枚举（0，0）就行了
          //for(i=0;i<n;i++)
            //for(j=0;j<m;j++)
             // dfs(i,j,0);
            printf("Scenario #%d:
",cas++);
          if(flag)
          {

               for(i=0;i<n*m;i++)
               printf("%c%d",pos[i].y+'A',pos[i].x+1);
               printf("
");

          }
           else
          printf("impossible
");
          printf("
");

     }
     return 0;
}