A Knight's Journey POJ 2488

Knight's Journey

Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

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

Sample Output

Scenario #1:
A1

Scenario #2:
impossible

Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4

 1 #include <stdio.h>
 2 #include <string.h>
 3 
 4 int main()
 5 {
 6     int T;
 7     int i,j,k,p,q,z,flg,u,v,cas=0;
 8     int x[100],y[100],t[100],d[10][10];
 9     int a[9]={0,-1,1,-2,2,-2,2,-1,1};
10     int b[9]={0,-2,-2,-1,-1,1,1,2,2};
11     
12     scanf("%d",&T);
13     while(T--)
14     {
15         memset(t,0,sizeof(t));
16         memset(d,0,sizeof(d));
17         scanf("%d %d",&p,&q);
18         i=1;z=0;
19         x[i]=1,y[i]=1,d[i][i]=1;
20         while(i>0)
21         {
22             flg=0;
23             for(k=t[i]+1;k<=8;k++)
24             {
25                 u=x[i]+a[k],v=y[i]+b[k];
26                 if(u>0 && u<=p && v>0 && v<=q && d[u][v]==0)
27                 {
28                     x[i+1]=u,y[i+1]=v,d[u][v]=1;
29                     t[i]=k;
30                     flg=1;
31                     break;
32                 }
33             }
34 
35             if(flg==1 && i==p*q-1)
36             {
37                 z=1;
38                 break;
39             }
40 
41             else if(flg==1)
42                 i++;
43             else
44             {
45                 t[i]=d[x[i]][y[i]]=0;
46                 i--;
47             }
48         }
49         if(p==1 && q==1)
50             z=1;
51         printf("Scenario #%d:
",++cas);
52         if(z==1)
53         {
54             for(i=1;i<=p*q;i++)
55             {
56                 printf("%c%d",'A'+y[i]-1,x[i]);
57             }
58             printf("
");
59         }
60         else
61         {
62             printf("impossible
");
63         }
64         printf("
");
65     }
66     return 0;
67 }

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