给定一个大小为N x N的方形棋盘，给出了骑士的位置和目标的位置，任务是找出骑士要达到目标位置所需的最小步骤。
例子 ：
Input : (2, 4) - knight's position, (6, 4) - target cell
Output : 2
Input : (4, 5) (1, 1)
Output : 3
在上一篇文章中已经讨论了解决上述问题的BFS方法。在这篇文章中，讨论了动态编程解决方案。
方法说明：
- 情况1：如果目标不在骑士位置的一行或一列上。
让一个8 x 8格的棋盘。现在，假设骑士在(3，3)，目标在(7，8)。从骑士的当前位置可能有8步移动，即(2，1)，(1、2)，(4、1)，(1、4)，(5、2)，(2、5)，(5 ，4)，(4，5)。但是，在这两个动作中，只有两个动作(5、4)和(4、5)会朝向目标，而其他所有动作都会远离目标。因此，要找到最小步数，请转到(4，5)或(5，4)。现在，计算从(4，5)和(5，4)达到目标所需的最小步骤。这是通过动态编程计算的。因此，这导致从(3，3)到(7，8)的最小步长。 - 情况2：如果目标是沿着骑士位置的一行或一列。
让一个8 x 8格的棋盘。现在，假设骑士在(4，3)，目标在(4，7)。可能有8个动作，但朝向目标，只有4个动作，即(5，5)，(3，5)，(2，4)，(6，4)。因为(5，5)等效于(3，5)，而(2，4)等效于(6，4)。因此，从这4个点可以将其转换为2个点。取(5，5)和(6，4)(在这里)。现在，计算从这两点到达到目标的最小步骤。这是通过动态编程计算的。因此，这导致从(4，3)到(4，7)的最小步长。
例外：当骑士将在拐角处并且目标与骑士位置的x和y坐标之差为(1，1)时，反之亦然。然后最少的步骤将是4。
动态规划方程：
1) dp[diffOfX][diffOfY] is the minimum steps taken from knight’s position to target’s position.
2) dp[diffOfX][diffOfY] = dp[diffOfY][diffOfX].
where, diffOfX = difference between knight’s x-coordinate and target’s x-coordinate
diffOfY = difference between knight’s y-coordinate and target’s y-coordinate
下面是上述方法的实现：
C++
// 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 than 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 than 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 than 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 PrinciRaj1992C#
// 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 than 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*/输出：
3