给定一个 N x N 大小的正方形棋盘，给定骑士的位置和目标的位置，任务是找出骑士到达目标位置所需的最小步数。
例子 ：
Input : (2, 4) - knight's position, (6, 4) - target cell
Output : 2
Input : (4, 5) (1, 1)
Output : 3
解决上述问题的 BFS 方法已经在上一篇文章中讨论过。在这篇文章中，讨论了动态规划解决方案。
方法说明：
- 情况1：如果目标不在骑士位置的一行或一列。
让一个 8 x 8 单元格的棋盘。现在，假设骑士在 (3, 3) 处，目标在 (7, 8) 处。从骑士的当前位置有可能的 8 个移动，即 (2, 1), (1, 2), (4, 1), (1, 4), (5, 2), (2, 5), (5) , 4), (4, 5)。但是，在这两个移动中，只有 (5, 4) 和 (4, 5) 将朝向目标，所有其他移动都远离目标。因此，要找到最小步骤，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数。这是通过动态规划计算的。因此，这导致从 (3, 3) 到 (7, 8) 的最小步骤。 - 情况2：如果目标沿着骑士位置的一行或一列。
让一个 8 x 8 单元格的棋盘。现在，假设骑士在 (4, 3) 处，目标在 (4, 7) 处。可能有 8 个移动，但是朝向目标，只有 4 个移动，即 (5, 5), (3, 5), (2, 4), (6, 4)。因为，(5, 5) 等价于(3, 5)，(2, 4) 等价于(6, 4)。所以，从这4点，可以换算成2点。取 (5, 5) 和 (6, 4) (这里)。现在，计算从这两个点到达目标所需的最小步数。这是通过动态规划计算的。因此，这导致从 (4, 3) 到 (4, 7) 的最小步骤。
例外：当骑士将在角落并且目标是这样的时，x 和 y 坐标与骑士的位置的差为 (1, 1)，反之亦然。那么最小步骤将是 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
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