Mindestschritte, um ein Ritter sein Ziel zu erreichen | Satz 2
Auf einem quadratischen Schachbrett der Größe N x N wird die Position des Ritters und die Position eines Ziels vorgegeben. Die Aufgabe besteht darin, die Mindestschritte herauszufinden, die ein Ritter unternehmen muss, um die Zielposition zu erreichen.
Beispiele:
Input : (2 4) - knight's position (6 4) - target cell Output : 2 Input : (4 5) (1 1) Output : 3
Ein BFS-Ansatz zur Lösung des oben genannten Problems wurde bereits im diskutiert vorherige Post. In diesem Beitrag wird eine dynamische Programmierlösung besprochen.
Erläuterung des Ansatzes:
Lassen Sie ein Schachbrett mit 8 x 8 Zellen. Nehmen wir nun an, der Springer steht bei (3 3) und das Ziel liegt bei (7 8). Von der aktuellen Position des Springers aus sind 8 Züge möglich, d. h. (2 1) (1 2) (4 1) (1 4) (5 2) (2 5) (5 4) (4 5). Aber von diesen werden nur zwei Bewegungen (5 4) und (4 5) in Richtung des Ziels erfolgen, während alle anderen vom Ziel weggehen. Um die Mindestschritte zu finden, gehen Sie also entweder zu (4 5) oder (5 4). Berechnen Sie nun die Mindestschritte aus (4 5) und (5 4), um das Ziel zu erreichen. Dies wird durch dynamische Programmierung berechnet. Somit ergeben sich die minimalen Schritte von (3 3) bis (7 8).
Lassen Sie ein Schachbrett mit 8 x 8 Zellen. Nehmen wir nun an, dass der Springer bei (4 3) steht und das Ziel bei (4 7) ist. Es sind 8 Züge möglich, aber in Richtung des Ziels gibt es nur 4 Züge, d. h. (5 5) (3 5) (2 4) (6 4). Da (5 5) äquivalent zu (3 5) und (2 4) äquivalent zu (6 4) ist. Aus diesen 4 Punkten kann also in 2 Punkte umgerechnet werden. Nimm (5 5) und (6 4) (hier). Berechnen Sie nun die Mindestschritte, die von diesen beiden Punkten aus unternommen werden müssen, um das Ziel zu erreichen. Dies wird durch dynamische Programmierung berechnet. Somit ergeben sich die minimalen Schritte von (4 3) bis (4 7).
Ausnahme : Wenn der Springer an der Ecke steht und das Ziel so ist, dass die Differenz der x- und y-Koordinaten mit der Position des Springers (1 1) beträgt oder umgekehrt. Dann betragen die Mindestschritte 4.
Dynamische Programmiergleichung:
1) dp[diffOfX][diffOfY] sind die minimalen Schritte, die von der Position des Springers zur Position des Ziels zurückgelegt werden müssen.
2) dp[diffOfX][diffOfY] = dp[diffOfY][diffOfX] .
wobei diffOfX = Differenz zwischen der X-Koordinate des Ritters und der X-Koordinate des Ziels
diffOfY = Differenz zwischen der Y-Koordinate des Ritters und der Y-Koordinate des Ziels
Nachfolgend finden Sie die Umsetzung des oben genannten Ansatzes:
// C++ code for minimum steps for // a knight to reach target position #include using namespace std ; // initializing the matrix. int dp [ 8 ][ 8 ] = { 0 }; int getsteps ( int x int y int tx int ty ) { // if knight is on the target // position return 0. if ( x == tx && y == ty ) return dp [ 0 ][ 0 ]; else { // if already calculated then return // that value. Taking absolute difference. if ( dp [ abs ( x - tx )][ abs ( y - ty )] != 0 ) return dp [ abs ( x - tx )][ abs ( y - ty )]; else { // there will be two distinct positions // from the knight towards a target. // if the target is in same row or column // as of knight then there can be four // positions towards the target but in that // two would be the same and the other two // would be the same. int x1 y1 x2 y2 ; // (x1 y1) and (x2 y2) are two positions. // these can be different according to situation. // From position of knight the chess board can be // divided into four blocks i.e.. N-E E-S S-W W-N . if ( x <= tx ) { if ( y <= ty ) { x1 = x + 2 ; y1 = y + 1 ; x2 = x + 1 ; y2 = y + 2 ; } else { x1 = x + 2 ; y1 = y - 1 ; x2 = x + 1 ; y2 = y - 2 ; } } else { if ( y <= ty ) { x1 = x - 2 ; y1 = y + 1 ; x2 = x - 1 ; y2 = y + 2 ; } else { x1 = x - 2 ; y1 = y - 1 ; x2 = x - 1 ; y2 = y - 2 ; } } // ans will be 1 + minimum of steps // required from (x1 y1) and (x2 y2). dp [ abs ( x - tx )][ abs ( y - ty )] = min ( getsteps ( x1 y1 tx ty ) getsteps ( x2 y2 tx ty )) + 1 ; // exchanging the coordinates x with y of both // knight and target will result in same ans. dp [ abs ( y - ty )][ abs ( x - tx )] = dp [ abs ( x - tx )][ abs ( y - ty )]; return dp [ abs ( x - tx )][ abs ( y - ty )]; } } } // Driver Code int main () { int i n x y tx ty ans ; // size of chess board n*n n = 100 ; // (x y) coordinate of the knight. // (tx ty) coordinate of the target position. x = 4 ; y = 5 ; tx = 1 ; ty = 1 ; // (Exception) these are the four corner points // for which the minimum steps is 4. if (( x == 1 && y == 1 && tx == 2 && ty == 2 ) || ( x == 2 && y == 2 && tx == 1 && ty == 1 )) ans = 4 ; else if (( x == 1 && y == n && tx == 2 && ty == n - 1 ) || ( x == 2 && y == n - 1 && tx == 1 && ty == n )) ans = 4 ; else if (( x == n && y == 1 && tx == n - 1 && ty == 2 ) || ( x == n - 1 && y == 2 && tx == n && ty == 1 )) ans = 4 ; else if (( x == n && y == n && tx == n - 1 && ty == n - 1 ) || ( x == n - 1 && y == n - 1 && tx == n && ty == n )) ans = 4 ; else { // dp[a][b] here a b is the difference of // x & tx and y & ty respectively. dp [ 1 ][ 0 ] = 3 ; dp [ 0 ][ 1 ] = 3 ; dp [ 1 ][ 1 ] = 2 ; dp [ 2 ][ 0 ] = 2 ; dp [ 0 ][ 2 ] = 2 ; dp [ 2 ][ 1 ] = 1 ; dp [ 1 ][ 2 ] = 1 ; ans = getsteps ( x y tx ty ); } cout < < ans < < endl ; return 0 ; }
Java //Java code for minimum steps for // a knight to reach target position public class GFG { // initializing the matrix. static int dp [][] = new int [ 8 ][ 8 ] ; static int getsteps ( int x int y int tx int ty ) { // if knight is on the target // position return 0. if ( x == tx && y == ty ) { return dp [ 0 ][ 0 ] ; } else // if already calculated then return // that value. Taking absolute difference. if ( dp [ Math . abs ( x - tx ) ][ Math . abs ( y - ty ) ] != 0 ) { return dp [ Math . abs ( x - tx ) ][ Math . abs ( y - ty ) ] ; } else { // there will be two distinct positions // from the knight towards a target. // if the target is in same row or column // as of knight then there can be four // positions towards the target but in that // two would be the same and the other two // would be the same. int x1 y1 x2 y2 ; // (x1 y1) and (x2 y2) are two positions. // these can be different according to situation. // From position of knight the chess board can be // divided into four blocks i.e.. N-E E-S S-W W-N . if ( x <= tx ) { if ( y <= ty ) { x1 = x + 2 ; y1 = y + 1 ; x2 = x + 1 ; y2 = y + 2 ; } else { x1 = x + 2 ; y1 = y - 1 ; x2 = x + 1 ; y2 = y - 2 ; } } else if ( y <= ty ) { x1 = x - 2 ; y1 = y + 1 ; x2 = x - 1 ; y2 = y + 2 ; } else { x1 = x - 2 ; y1 = y - 1 ; x2 = x - 1 ; y2 = y - 2 ; } // ans will be 1 + minimum of steps // required from (x1 y1) and (x2 y2). dp [ Math . abs ( x - tx ) ][ Math . abs ( y - ty ) ] = Math . min ( getsteps ( x1 y1 tx ty ) getsteps ( x2 y2 tx ty )) + 1 ; // exchanging the coordinates x with y of both // knight and target will result in same ans. dp [ Math . abs ( y - ty ) ][ Math . abs ( x - tx ) ] = dp [ Math . abs ( x - tx ) ][ Math . abs ( y - ty ) ] ; return dp [ Math . abs ( x - tx ) ][ Math . abs ( y - ty ) ] ; } } // Driver Code static public void main ( String [] args ) { int i n x y tx ty ans ; // size of chess board n*n n = 100 ; // (x y) coordinate of the knight. // (tx ty) coordinate of the target position. x = 4 ; y = 5 ; tx = 1 ; ty = 1 ; // (Exception) these are the four corner points // for which the minimum steps is 4. if (( x == 1 && y == 1 && tx == 2 && ty == 2 ) || ( x == 2 && y == 2 && tx == 1 && ty == 1 )) { ans = 4 ; } else if (( x == 1 && y == n && tx == 2 && ty == n - 1 ) || ( x == 2 && y == n - 1 && tx == 1 && ty == n )) { ans = 4 ; } else if (( x == n && y == 1 && tx == n - 1 && ty == 2 ) || ( x == n - 1 && y == 2 && tx == n && ty == 1 )) { ans = 4 ; } else if (( x == n && y == n && tx == n - 1 && ty == n - 1 ) || ( x == n - 1 && y == n - 1 && tx == n && ty == n )) { ans = 4 ; } else { // dp[a][b] here a b is the difference of // x & tx and y & ty respectively. dp [ 1 ][ 0 ] = 3 ; dp [ 0 ][ 1 ] = 3 ; dp [ 1 ][ 1 ] = 2 ; dp [ 2 ][ 0 ] = 2 ; dp [ 0 ][ 2 ] = 2 ; dp [ 2 ][ 1 ] = 1 ; dp [ 1 ][ 2 ] = 1 ; ans = getsteps ( x y tx ty ); } System . out . println ( ans ); } } /*This code is contributed by PrinciRaj1992*/
Python3 # Python3 code for minimum steps for # a knight to reach target position # initializing the matrix. dp = [[ 0 for i in range ( 8 )] for j in range ( 8 )]; def getsteps ( x y tx ty ): # if knight is on the target # position return 0. if ( x == tx and y == ty ): return dp [ 0 ][ 0 ]; # if already calculated then return # that value. Taking absolute difference. elif ( dp [ abs ( x - tx )][ abs ( y - ty )] != 0 ): return dp [ abs ( x - tx )][ abs ( y - ty )]; else : # there will be two distinct positions # from the knight towards a target. # if the target is in same row or column # as of knight then there can be four # positions towards the target but in that # two would be the same and the other two # would be the same. x1 y1 x2 y2 = 0 0 0 0 ; # (x1 y1) and (x2 y2) are two positions. # these can be different according to situation. # From position of knight the chess board can be # divided into four blocks i.e.. N-E E-S S-W W-N . if ( x <= tx ): if ( y <= ty ): x1 = x + 2 ; y1 = y + 1 ; x2 = x + 1 ; y2 = y + 2 ; else : x1 = x + 2 ; y1 = y - 1 ; x2 = x + 1 ; y2 = y - 2 ; elif ( y <= ty ): x1 = x - 2 ; y1 = y + 1 ; x2 = x - 1 ; y2 = y + 2 ; else : x1 = x - 2 ; y1 = y - 1 ; x2 = x - 1 ; y2 = y - 2 ; # ans will be 1 + minimum of steps # required from (x1 y1) and (x2 y2). dp [ abs ( x - tx )][ abs ( y - ty )] = min ( getsteps ( x1 y1 tx ty ) getsteps ( x2 y2 tx ty )) + 1 ; # exchanging the coordinates x with y of both # knight and target will result in same ans. dp [ abs ( y - ty )][ abs ( x - tx )] = dp [ abs ( x - tx )][ abs ( y - ty )]; return dp [ abs ( x - tx )][ abs ( y - ty )]; # Driver Code if __name__ == '__main__' : # size of chess board n*n n = 100 ; # (x y) coordinate of the knight. # (tx ty) coordinate of the target position. x = 4 ; y = 5 ; tx = 1 ; ty = 1 ; # (Exception) these are the four corner points # for which the minimum steps is 4. if (( x == 1 and y == 1 and tx == 2 and ty == 2 ) or ( x == 2 and y == 2 and tx == 1 and ty == 1 )): ans = 4 ; elif (( x == 1 and y == n and tx == 2 and ty == n - 1 ) or ( x == 2 and y == n - 1 and tx == 1 and ty == n )): ans = 4 ; elif (( x == n and y == 1 and tx == n - 1 and ty == 2 ) or ( x == n - 1 and y == 2 and tx == n and ty == 1 )): ans = 4 ; elif (( x == n and y == n and tx == n - 1 and ty == n - 1 ) or ( x == n - 1 and y == n - 1 and tx == n and ty == n )): ans = 4 ; else : # dp[a][b] here a b is the difference of # x & tx and y & ty respectively. dp [ 1 ][ 0 ] = 3 ; dp [ 0 ][ 1 ] = 3 ; dp [ 1 ][ 1 ] = 2 ; dp [ 2 ][ 0 ] = 2 ; dp [ 0 ][ 2 ] = 2 ; dp [ 2 ][ 1 ] = 1 ; dp [ 1 ][ 2 ] = 1 ; ans = getsteps ( x y tx ty ); print ( ans ); # This code is contributed by PrinciRaj1992
C# // C# code for minimum steps for // a knight to reach target position using System ; public class GFG { // initializing the matrix. static int [ ] dp = new int [ 8 8 ]; static int getsteps ( int x int y int tx int ty ) { // if knight is on the target // position return 0. if ( x == tx && y == ty ) { return dp [ 0 0 ]; } else // if already calculated then return // that value. Taking Absolute difference. if ( dp [ Math . Abs ( x - tx ) Math . Abs ( y - ty )] != 0 ) { return dp [ Math . Abs ( x - tx ) Math . Abs ( y - ty )]; } else { // there will be two distinct positions // from the knight towards a target. // if the target is in same row or column // as of knight then there can be four // positions towards the target but in that // two would be the same and the other two // would be the same. int x1 y1 x2 y2 ; // (x1 y1) and (x2 y2) are two positions. // these can be different according to situation. // From position of knight the chess board can be // divided into four blocks i.e.. N-E E-S S-W W-N . if ( x <= tx ) { if ( y <= ty ) { x1 = x + 2 ; y1 = y + 1 ; x2 = x + 1 ; y2 = y + 2 ; } else { x1 = x + 2 ; y1 = y - 1 ; x2 = x + 1 ; y2 = y - 2 ; } } else if ( y <= ty ) { x1 = x - 2 ; y1 = y + 1 ; x2 = x - 1 ; y2 = y + 2 ; } else { x1 = x - 2 ; y1 = y - 1 ; x2 = x - 1 ; y2 = y - 2 ; } // ans will be 1 + minimum of steps // required from (x1 y1) and (x2 y2). dp [ Math . Abs ( x - tx ) Math . Abs ( y - ty )] = Math . Min ( getsteps ( x1 y1 tx ty ) getsteps ( x2 y2 tx ty )) + 1 ; // exchanging the coordinates x with y of both // knight and target will result in same ans. dp [ Math . Abs ( y - ty ) Math . Abs ( x - tx )] = dp [ Math . Abs ( x - tx ) Math . Abs ( y - ty )]; return dp [ Math . Abs ( x - tx ) Math . Abs ( y - ty )]; } } // Driver Code static public void Main () { int i n x y tx ty ans ; // size of chess board n*n n = 100 ; // (x y) coordinate of the knight. // (tx ty) coordinate of the target position. x = 4 ; y = 5 ; tx = 1 ; ty = 1 ; // (Exception) these are the four corner points // for which the minimum steps is 4. if (( x == 1 && y == 1 && tx == 2 && ty == 2 ) || ( x == 2 && y == 2 && tx == 1 && ty == 1 )) { ans = 4 ; } else if (( x == 1 && y == n && tx == 2 && ty == n - 1 ) || ( x == 2 && y == n - 1 && tx == 1 && ty == n )) { ans = 4 ; } else if (( x == n && y == 1 && tx == n - 1 && ty == 2 ) || ( x == n - 1 && y == 2 && tx == n && ty == 1 )) { ans = 4 ; } else if (( x == n && y == n && tx == n - 1 && ty == n - 1 ) || ( x == n - 1 && y == n - 1 && tx == n && ty == n )) { ans = 4 ; } else { // dp[a b] here a b is the difference of // x & tx and y & ty respectively. dp [ 1 0 ] = 3 ; dp [ 0 1 ] = 3 ; dp [ 1 1 ] = 2 ; dp [ 2 0 ] = 2 ; dp [ 0 2 ] = 2 ; dp [ 2 1 ] = 1 ; dp [ 1 2 ] = 1 ; ans = getsteps ( x y tx ty ); } Console . WriteLine ( ans ); } } /*This code is contributed by PrinciRaj1992*/
JavaScript < script > // JavaScript code for minimum steps for // a knight to reach target position // initializing the matrix. let dp = new Array ( 8 ) for ( let i = 0 ; i < 8 ; i ++ ){ dp [ i ] = new Array ( 8 ). fill ( 0 ) } function getsteps ( x y tx ty ) { // if knight is on the target // position return 0. if ( x == tx && y == ty ) return dp [ 0 ][ 0 ]; else { // if already calculated then return // that value. Taking absolute difference. if ( dp [( Math . abs ( x - tx ))][( Math . abs ( y - ty ))] != 0 ) return dp [( Math . abs ( x - tx ))][( Math . abs ( y - ty ))]; else { // there will be two distinct positions // from the knight towards a target. // if the target is in same row or column // as of knight then there can be four // positions towards the target but in that // two would be the same and the other two // would be the same. let x1 y1 x2 y2 ; // (x1 y1) and (x2 y2) are two positions. // these can be different according to situation. // From position of knight the chess board can be // divided into four blocks i.e.. N-E E-S S-W W-N . if ( x <= tx ) { if ( y <= ty ) { x1 = x + 2 ; y1 = y + 1 ; x2 = x + 1 ; y2 = y + 2 ; } else { x1 = x + 2 ; y1 = y - 1 ; x2 = x + 1 ; y2 = y - 2 ; } } else { if ( y <= ty ) { x1 = x - 2 ; y1 = y + 1 ; x2 = x - 1 ; y2 = y + 2 ; } else { x1 = x - 2 ; y1 = y - 1 ; x2 = x - 1 ; y2 = y - 2 ; } } // ans will be 1 + minimum of steps // required from (x1 y1) and (x2 y2). dp [( Math . abs ( x - tx ))][( Math . abs ( y - ty ))] = Math . min ( getsteps ( x1 y1 tx ty ) getsteps ( x2 y2 tx ty )) + 1 ; // exchanging the coordinates x with y of both // knight and target will result in same ans. dp [( Math . abs ( y - ty ))][( Math . abs ( x - tx ))] = dp [( Math . abs ( x - tx ))][( Math . abs ( y - ty ))]; return dp [( Math . abs ( x - tx ))][( Math . abs ( y - ty ))]; } } } // Driver Code let i n x y tx ty ans ; // size of chess board n*n n = 100 ; // (x y) coordinate of the knight. // (tx ty) coordinate of the target position. x = 4 ; y = 5 ; tx = 1 ; ty = 1 ; // (Exception) these are the four corner points // for which the minimum steps is 4. if (( x == 1 && y == 1 && tx == 2 && ty == 2 ) || ( x == 2 && y == 2 && tx == 1 && ty == 1 )) ans = 4 ; else if (( x == 1 && y == n && tx == 2 && ty == n - 1 ) || ( x == 2 && y == n - 1 && tx == 1 && ty == n )) ans = 4 ; else if (( x == n && y == 1 && tx == n - 1 && ty == 2 ) || ( x == n - 1 && y == 2 && tx == n && ty == 1 )) ans = 4 ; else if (( x == n && y == n && tx == n - 1 && ty == n - 1 ) || ( x == n - 1 && y == n - 1 && tx == n && ty == n )) ans = 4 ; else { // dp[a][b] here a b is the difference of // x & tx and y & ty respectively. dp [ 1 ][ 0 ] = 3 ; dp [ 0 ][ 1 ] = 3 ; dp [ 1 ][ 1 ] = 2 ; dp [ 2 ][ 0 ] = 2 ; dp [ 0 ][ 2 ] = 2 ; dp [ 2 ][ 1 ] = 1 ; dp [ 1 ][ 2 ] = 1 ; ans = getsteps ( x y tx ty ); } document . write ( ans ' ' ); // This code is contributed by shinjanpatra. < /script>
Ausgabe:
3
Zeitkomplexität: O(N * M) wobei N die Gesamtzahl der Zeilen und M die Gesamtzahl der Spalten ist
Hilfsraum: O(N * M)
Das Könnte Ihnen Gefallen
Top Artikel
