Дадена е квадратна шахматна дъска с размер N x N, дадена е позицията на коня и позицията на целта. Задачата е да откриете минималните стъпки, които конят ще предприеме, за да достигне целевата позиция.
Примери:
Input : (2 4) - knight's position (6 4) - target cell Output : 2 Input : (4 5) (1 1) Output : 3
Подход на BFS за решаване на горния проблем вече беше обсъден в предишен пост. В тази публикация се обсъжда решение за динамично програмиране.
Обяснение на подхода:
Нека шахматна дъска от 8 x 8 клетки. Сега да кажем, че рицарят е на (3 3), а целта е на (7 8). Има възможни 8 хода от текущата позиция на коня, т.е. (2 1) (1 2) (4 1) (1 4) (5 2) (2 5) (5 4) (4 5). Но сред тези само два хода (5 4) и (4 5) ще бъдат към целта, а всички останали се отдалечават от целта. Така че за намиране на минимални стъпки отидете на (4 5) или (5 4). Сега изчислете минималните стъпки, направени от (4 5) и (5 4), за да достигнете целта. Това се изчислява чрез динамично програмиране. Така това води до минималните стъпки от (3 3) до (7 8).
Нека шахматна дъска от 8 x 8 клетки. Сега да кажем, че рицарят е на (4 3), а целта е на (4 7). Има възможни 8 хода, но към целта има само 4 хода, т.е. (5 5) (3 5) (2 4) (6 4). Тъй като (5 5) е еквивалентно на (3 5) и (2 4) е еквивалентно на (6 4). Така че от тези 4 точки може да се преобразува в 2 точки. Вземане на (5 5) и (6 4) (тук). Сега изчислете минималните стъпки, направени от тези две точки, за да достигнете целта. Това се изчислява чрез динамично програмиране. Така това води до минималните стъпки от (4 3) до (4 7).
Изключение: Когато рицарят ще бъде в ъгъла и целта е такава, че разликата на координатите x и y с позицията на рицаря е (1 1) или обратното. Тогава минималните стъпки ще бъдат 4.
Уравнение за динамично програмиране:
1) dp[diffOfX][diffOfY] е минималният брой стъпки, направени от позицията на коня до позицията на целта.
2) dp[diffOfX][diffOfY] = dp[diffOfY][diffOfX] .
където diffOfX = разликата между х-координатата на коня и х-координатата на целта
diffOfY = разлика между y-координатата на коня и y-координатата на целта
По-долу е изпълнението на горния подход:
// C++ code for minimum steps for // a knight to reach target position #include
//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 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# 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*/
<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(xytxty) { // 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>
Изход:
3
Времева сложност: O(N * M), където N е общият брой редове, а M е общият брой колони
Помощно пространство: O(N * M)