C++ :

#include <bits/stdc++.h>
using namespace std;
//棋盘大小及起止位置 
int n,m,s1,s2,t1,t2;
int a[10][10];//存放棋盘
int d[10][10];//到每个位置的最短路径 

//dep：到该点的步数 
void fun(int x,int y,int dep){
	//如果走到这个位置的步数比假设的少（也可以防止死循环） 
	if(dep < d[x][y]){
		d[x][y] = dep;
		//八个方向探测
		if(x-1>=1&&y-2>=1) fun(x-1,y-2,dep+1); 
		if(x-2>=1&&y-1>=1) fun(x-2,y-1,dep+1);
		if(x-2>=1&&y+1<=m) fun(x-2,y+1,dep+1);
		if(x-1>=1&&y+2<=m) fun(x-1,y+2,dep+1);
		if(x+1<=n&&y+2<=m) fun(x+1,y+2,dep+1);
		if(x+2<=n&&y+1<=m) fun(x+2,y+1,dep+1);
		if(x+2<=n&&y-1>=1) fun(x+2,y-1,dep+1);
		if(x+1<=n&&y-2>=1) fun(x+1,y-2,dep+1); 
	}	
}

int main(){
	cin>>n>>m>>s1>>s2>>t1>>t2;
	for(int i = 1;i <= n;i++){
		for(int j = 1;j <= m;j++){
			d[i][j] = INT_MAX;//先设置为最大 
		}
	}
	fun(s1,s2,0);
	cout<<d[t1][t2]<<endl;
}

Java :

import java.util.HashMap;
import java.util.LinkedList;
import java.util.Map;
import java.util.Queue;
import java.util.Scanner;

public class Main {
	static int n;
	static int m;
	static int x;
	static int y;
	static int s;
	static int t;
	//记录步数
	static int a[][] = new int[6][6];
	//方向数组
	static int fx[] = {-1,-2,-2,-1,1,2,2,1};
	static int fy[] = {-2,-1,1,2,2,1,-1,-2};
	//队列
	static Queue<Map<String,Integer>> Q = new LinkedList<Map<String,Integer>>();
	
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		n = sc.nextInt();
		m = sc.nextInt();
		x = sc.nextInt();
		y = sc.nextInt();
		s = sc.nextInt();
		t = sc.nextInt();
		//地图所有格子都是最大值
		int i,j,oi,oj,od;
		for(i=1;i<=n;i++) {
			for(j=1;j<=m;j++) {
				a[i][j] = Integer.MAX_VALUE;
			}
		}
		a[x][y] = 0;
		//now记录当前元素  next下一个元素
		Map<String,Integer> now = new HashMap<String,Integer>();
		Map<String,Integer> next = null;

		//起始点入队
		now.put("i",x);
		now.put("j",y);
		now.put("d",0);
		Q.add(now);
		//如果队列有元素
		while(Q.isEmpty()==false) {
			now = new HashMap<String,Integer>();
			now = Q.peek();
			Q.remove();
			//必须符合要求才能入队
			for(int k=0;k<8;k++) {
				next = new HashMap<String,Integer>();
				next.put("i", now.get("i")+fx[k]);
				next.put("j", now.get("j")+fy[k]);
				next.put("d", now.get("d")+1);
				oi = next.get("i");
				oj = next.get("j");
				od = next.get("d");
				if(oi>=1 && oi<=n && oj>=1 && oj<=m && od < a[oi][oj]) {
					//说明next这个点就是下一个入队点
					a[oi][oj] = od;
					Q.add(next);
				}
			}
		}
		System.out.println(a[s][t]);
		
		sc.close();
	}
}