POJ 2488, A Knight's Journey

Time Limit: 1000MS  Memory Limit: 65536K
Total Submissions: 7832  Accepted: 2671

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

 

Source
TUD Programming Contest 2005, Darmstadt, Germany

---

//  POJ2488.cpp : Defines the entry point for the console application.
//

#include  < iostream >
using   namespace  std;

static   int  step[ 8 ][ 2 ]  =  { - 2 ,  - 1 ,  - 2 ,  1 ,  - 1 ,  - 2 ,  - 1 ,  2 ,  1 ,  - 2 ,  1 ,  2 ,  2 ,  - 1 ,  2 ,  1 };
bool  DFS( int  x,  int  y,  int  p,  int  q,  bool  board[ 26 ][ 26 ],  char  path[ 54 ],  int  marked)
{
     if  (marked  ==  p  *  q)  return   true ;
     int  x1, y1;
     for  ( int  i  =   0 ; i  <   8 ;  ++ i)
    {
        x1  =  x  +  step[i][ 0 ];
        y1  =  y  +  step[i][ 1 ];
         if  (x1  >=   0   &&  x1  <  q  &&  y1  >=   0   &&  y1  <  p  &&  board[y1][x1]  ==   false )
        {
            board[y1][x1]  =   true ;
            path[(marked << 1 )]  =  x1  +   ' A ' ;
            path[(marked << 1 )  +   1 ]  =  y1  +   ' 1 ' ;
             if (DFS(x1,y1,p,q,board,path,marked  +   1 ))  return   true ;
            board[y1][x1]  =   false ;
        }
    }
     return   false ;
}

int  main( int  argc,  char *  argv[])
{
     bool  board[ 26 ][ 26 ];
     char  path[ 54 ];
     int  cases;
    cin  >>  cases;
     for  ( int  c  =   1 ; c  <=  cases;  ++ c)
    {
         int  p, q;
        cin  >>  p  >>  q;
        memset(board,  0 , sizeof (board));
        memset(path,  0 , sizeof (path));

        board[ 0 ][ 0 ]  =   true ;
        path[ 0 ]  =   ' A ' ;
        path[ 1 ]  =   ' 1 ' ;
         if  (DFS( 0 , 0 ,p,q,board,path, 1 ))
            cout  <<   " Scenario # " << c << " :\n " << path << endl << endl;
         else
            cout  <<   " Scenario # " << c << " :\nimpossible\n " << endl;
    };
     return   0 ;
}

转载于:https://www.cnblogs.com/asuran/archive/2009/10/10/1579985.html