题目链接：过河卒

初始条件： f(0,1) = f(1,0) =1
关系式：f(i,j) = f(i-1,j) + f(i,j-1)

java代码：

import java.util.Arrays;
import java.util.Scanner;

public class Main {
	
	public static void main(String[] args) {
		
		int n,m,a,b;
		Scanner in = new Scanner(System.in);
//		坐标加2，可以避免：在标记不能走的位置时发生数组越界的情况
//		简单来说就是让左边和上边空出来两层不使用
		n = in.nextInt()+2;
		m = in.nextInt()+2;
		a = in.nextInt()+2;
		b = in.nextInt()+2;
		
//		初始化数组
		long arr1[][] = new long[n+1][m+1];
		boolean arr2[][] = new boolean[n+1][m+1];
		for(int i=0;i<=n;++i) {
			Arrays.fill(arr1[i], 0);
			Arrays.fill(arr2[i], true);
		}
		
//		标记不能走的位置
		arr2[a][b] = false;
		arr2[a-1][b-2] = false;
		arr2[a-2][b-1] = false;
		arr2[a-2][b+1] = false;
		arr2[a-1][b+2] = false;
		arr2[a+1][b-2] = false;
		arr2[a+2][b-1] = false;
		arr2[a+1][b+2] = false;
		arr2[a+2][b+1] = false;
		
//		开始递推计算
		arr1[2][2] = 1;
		for(int i=2;i<=n;++i) {
			for(int j=2;j<=m;++j) {
				if(i==2&&j==2) continue;
				if(arr2[i][j]!=false)
					arr1[i][j] = arr1[i-1][j]+arr1[i][j-1];
			}
		}
		
		System.out.println(arr1[n][m]);
	}
}