Description
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.
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
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 }