洛谷p1002过河卒java动态规划

洛谷p1002过河卒是一道动态规划问题，题目要求在一个棋盘上，卒从起点出发，不能经过马的控制点，到达终点的所有路径数。以下是Java实现的动态规划解法：

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int x1 = sc.nextInt();
        int y1 = sc.nextInt();
        int x2 = sc.nextInt();
        int y2 = sc.nextInt();
        int[][] horse = new int[22][22];
        int[][] dp = new int[22][22];
        horse[x1][y1] = 1;
        dp[x1][y1] = 1;
        if (x1 - 1 >= 0 && y1 - 2 >= 0) horse[x1 - 1][y1 - 2] = 1;
        if (x1 - 2 >= 0 && y1 - 1 >= 0) horse[x1 - 2][y1 - 1] = 1;
        if (x1 - 2 >= 0 && y1 + 1 <= 20) horse[x1 - 2][y1 + 1] = 1;
        if (x1 - 1 >= 0 && y1 + 2 <= 20) horse[x1 - 1][y1 + 2] = 1;
        if (x1 + 1 <= 20 && y1 + 2 <= 20) horse[x1 + 1][y1 + 2] = 1;
        if (x1 + 2 <= 20 && y1 + 1 <= 20) horse[x1 + 2][y1 + 1] = 1;
        if (x1 + 2 <= 20 && y1 - 1 >= 0) horse[x1 + 2][y1 - 1] = 1;
        if (x1 + 1 <= 20 && y1 - 2 >= 0) horse[x1 + 1][y1 - 2] = 1;
        for (int i = x1 + 1; i <= x2; i++) {
            for (int j = y1; j <= y2; j++) {
                if (horse[i][j] == 0) {
                    dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
                }
            }
        }
        System.out.println(dp[x2][y2]);
    }
}