原題：

A friend of you is doing research on the Traveling Knight Problem (TKP) where you are to find the shortest closed tour of knight moves that visits each square of a given set of n squares on a chessboard exactly once. He thinks that the most difficult part of the problem is determining the smallest number of knight moves between two given squares and that, once you have accomplished this, finding the tour would be easy. 
Of course you know that it is vice versa. So you offer him to write a program that solves the "difficult" part. 

Your job is to write a program that takes two squares a and b as input and then determines the number of knight moves on a shortest route from a to b

題意：

和上一個象棋的題目一樣，只不過上一個要求走遍棋盤的每一個格子，這個要求走到一個點，基本一樣。

題解：

和上一個一樣的深搜。

程式碼：AC

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#define INF 100000+10
using namespace std;

int dx[8] = {-2,-2,2,2,-1,-1,1,1};
int dy[8] = {1,-1,-1,1,2,-2,-2,2};
int ans = 0,stx,sty,endx,endy;
int board[10][10];
int min(int a,int b)
{
	return a<b?a:b;
}
void dfs(int x,int y,int step)
{
	if(x < 0 || y < 0||x >= 8 ||y >= 8) return;
	if(x == endx&&y == endy)
	{
		ans = min(ans,step);
		return ;
	}
	if(step > 6) return;
	if(step > board[x][y]) return ;
	board[x][y] = step;
	int tx,ty;
	for(int i = 0;i <= 7;i++)
	{
		tx = x+dx[i];
		ty = y+dy[i];
		dfs(tx,ty,step+1);
	}
}
void init()
{  
    int i,j;  
    for(i=0;i<8;++i){  
        for(j=0;j<8;++j){  
            board[i][j] = INF;  
        }  
    }  
}  
int main()
{
	char v[2],n[2]; 
	while(scanf("%s %s",v,n) != EOF)
	{
		ans = INF;
		stx = v[0] - 'a';
		sty = v[1] - '0'-1;
		endx = n[0] - 'a';
		endy = n[1] - '0'-1;
		init();
		dfs(stx,sty,0);
		printf("To get from %s to %s takes %d knight moves.\n",v,n,ans);  
 }
    return 0;
}