Minsta steg för att nå målet av en riddare | Set 2
Givet ett fyrkantigt schackbräde av storleken N x N ges riddarens position och målets position uppgiften att ta reda på de minsta steg en riddare kommer att ta för att nå målpositionen.
Exempel:
Input : (2 4) - knight's position (6 4) - target cell Output : 2 Input : (4 5) (1 1) Output : 3
En BFS-metod för att lösa ovanstående problem har redan diskuterats i tidigare posta. I detta inlägg diskuteras en dynamisk programmeringslösning.
Förklaring av tillvägagångssättet:
Låt ett schackbräde på 8 x 8 celler. Låt nu säga att riddaren är på (3 3) och målet är på (7 8). Det finns 8 möjliga drag från riddarens nuvarande position, dvs (2 1) (1 2) (4 1) (1 4) (5 2) (2 5) (5 4) (4 5). Men bland dessa kommer endast två drag (5 4) och (4 5) att vara mot målet och alla andra går bort från målet. Så för att hitta minimisteg gå till antingen (4 5) eller (5 4). Beräkna nu de minsta stegen från (4 5) och (5 4) för att nå målet. Detta beräknas genom dynamisk programmering. Detta resulterar således i minimisteg från (3 3) till (7 8).
Låt ett schackbräde på 8 x 8 celler. Låt oss nu säga att riddaren är på (4 3) och målet är på (4 7). Det finns 8 möjliga drag men mot målet finns det bara 4 drag, dvs (5 5) (3 5) (2 4) (6 4). Som (5 5) är ekvivalent med (3 5) och (2 4) är ekvivalent med (6 4). Så från dessa 4 poäng kan det omvandlas till 2 poäng. Ta (5 5) och (6 4) (här). Beräkna nu de minsta steg som tas från dessa två punkter för att nå målet. Detta beräknas genom dynamisk programmering. Detta resulterar således i minimisteg från (4 3) till (4 7).
Undantag: När riddaren kommer att vara i hörnet och målet är sådan att skillnaden mellan x och y koordinater med riddarens position är (1 1) eller vice versa. Då blir minsta steg 4.
Dynamisk programmeringsekvation:
1) dp[diffOfX][diffOfY] är de minsta steg som tas från riddarens position till målets position.
2) dp[diffOfX][diffOfY] = dp[diffOfY][diffOfX] .
där diffOfX = skillnaden mellan riddarens x-koordinat och målets x-koordinat
diffOfY = skillnaden mellan riddarens y-koordinat och målets y-koordinat
Nedan är implementeringen av ovanstående tillvägagångssätt:
// 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>
Produktion:
3
Tidskomplexitet: O(N * M) där N är det totala antalet rader och M är det totala antalet kolumner
Hjälputrymme: O(N * M)
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