acm algorithm practice Jan 3 DFS

poj 2488 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.

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simple DFS problem, the key is "trace back"

pseudocode for DFS with traceback:

bool dfs(point p)

{

　　if (the count == total squares)

　　{

　　　　print and return true;

　　}

　　// 8 moving method

　　for each moving method 

　　　　if the p' is white && not exceed the boundary

　　　　　　p'.color <----- grey

　　　　　　count ++

　　　　　　inqueue(p')

　　　　　　if (dfs(p') == true)   // which means  if the the bottom recursion return the true, no need to traceback 

　　　　　　　　return true 

　　　　　　p'.color <----white  //trace back 

　　　　　　count --

　　　　　　dequeue(p')

 

　　return false

}

more convient way is to use recursion times replace count, it is a traditional recursive problem. 

bool dfs( int dip, int x, int y )
{
 if( dip == n )
 {
    int j;
    for( j = 0; j<dip; j++ )//找到答案则输出路径
    {
     printf( "%c%d", path[j].y+'A', path[j].x+1 );
    }
    cout << endl;
    return true;
 }
 int i;
 for( i = 0; i<8; i++ )
 {
    int tx, ty;
    tx = x + a_x[i];
    ty = y + a_y[i];
    if( !isover(tx, ty) && !f[ty][tx] )
    {
     f[ty][tx] = true;
     path[dip].x = tx;
     path[dip].y = ty;
     if( dfs( dip+1, tx, ty ) )
      return true;
     f[ty][tx] = false;//算是回朔吧
    }
 }
 return false;
}

here, since dip stands for the recursion times. and we use path[] record the path. it is not necessary to delete the value in path[] in traceback stage (the new data will overwrite the old error one)