Knight Moves Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4813    Accepted Submission(s): 2957


Problem Description
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.
 

Input
The input file will contain one or more test cases. Each test case consists of one line containing two squares separated by one space. A square is a string consisting of a letter (a-h) representing the column and a digit (1-8) representing the row on the chessboard.
 

Output
For each test case, print one line saying "To get from xx to yy takes n knight moves.".
 

Sample Input
e2 e4 a1 b2 b2 c3 a1 h8 a1 h7 h8 a1 b1 c3 f6 f6
 

Sample Output
To get from e2 to e4 takes 2 knight moves. To get from a1 to b2 takes 4 knight moves. To get from b2 to c3 takes 2 knight moves. To get from a1 to h8 takes 6 knight moves. To get from a1 to h7 takes 5 knight moves. To get from h8 to a1 takes 6 knight moves. To get from b1 to c3 takes 1 knight moves. To get from f6 to f6 takes 0 knight moves.
 

Source
University of Ulm Local Contest 1996
 

Recommend
Eddy
 
解题思路：搜索方式不同往常的广度优先搜索，只要搞清楚方向，直接搜索即可，其他都交给计算机，你只要输条件和等结果就可以了。
由于是走‘日’字型的，则有，若当前处于中间点，则有8个方向可以走。如图，若处于‘0’点，则有8个方向可以走（‘1’点）。



#include<cstdio> #include<cstring> #include<queue> using namespace std; int map[9][9]; int n=8; int dir[8][2]={1,2,1,-2,-1,2,-1,-2,2,1,2,-1,-2,1,-2,-1};     //八个方向哦 char a[2][2]; int sx,sy,ex,ey; struct node {   int x;   int y;   int step; }; bool inmap(int x,int y) {     if(x>=0&&x<n&&y>=0&&y<n)         return true;     return false; } int bfs() {     node s,e;     queue<node> q;     s.x=sx;     s.y=sy;     s.step=0;     map[sx][sy]=1;     q.push(s);     if(sx==ex&&sy==ey)         return 0;     while(!q.empty())     {         s=q.front();         q.pop();         for(int i=0;i<8;i++)     //直接走就对了         {             e.x=s.x+dir[i][0];             e.y=s.y+dir[i][1];             e.step=s.step+1;             if(inmap(e.x,e.y)&&!map[e.x][e.y])             {                 if(e.x==ex&&e.y==ey)                     return e.step;                 map[e.x][e.y]=1;                 q.push(e);             }         }     }     return -1; } int main() {     while(scanf("%s%s",a[0],a[1])!=EOF)     {         memset(map,0,sizeof(map));         sx=a[0][0]-'a';     //存储形式的转换，利于数组运用         sy=a[0][1]-'1';         ex=a[1][0]-'a';         ey=a[1][1]-'1';         printf("To get from %c%c to %c%c takes %d knight moves.\n",a[0][0],a[0][1],a[1][0],a[1][1],bfs());     }     return 0; }