Knight Moves ---BFS模板

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. 


你的一个朋友正在研究骑士旅行问题，在这问题中，你要找到最短骑士之旅移动步数，在象棋板上，被给设定的n个正方形数，仅能一次经过每个正方形，他认为，这个问题最困难的部分是决定骑士在两个正方形之间移动的最小次数，并且，一旦你完成这个，找到旅途是容易的。

当然，你知道它是反之亦然的，因此，你得写个程序给他去解决这个难点。

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. 
 a和b两个正方形是输入的，决定骑士移动最短路线从a到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.

#include<iostream>  #include<queue>  #include<cstdio>  using namespace std;  struct node  {      int c;      int r;      int lev;  } front,tmp,start,End;  queue <node>Q;  int a[]= {0,-2,-2,-1,-1,1,1,2,2};  int b[]= {0,-1,1,-2,2,-2,2,-1,1};  int GetRe(node s,node e)  {      int i;      while(!Q.empty())          Q.pop();      if(e.c==s.c&&e.r==s.r)return 0;      Q.push(s);      while(!Q.empty())      {          front=Q.front();          Q.pop();          for(i=1; i<=8; i++)          {              tmp.c=b[i]+front.c;              tmp.r=a[i]+front.r;              tmp.lev=front.lev+1;              if(e.c==tmp.c&&e.r==tmp.r)                  return tmp.lev;              if(tmp.c>=1&&tmp.c<=8&&tmp.r>=1&&tmp.r<=8)                  Q.push(tmp);          }      }  }  int main()  {      char s[3],e[3];      while(scanf("%s%s",s,e)!=EOF)      {          start.c=s[0]-'a'+1;          start.r=s[1]-'0';          start.lev=0;          End.c=e[0]-'a'+1;          End.r=e[1]-'0';          printf("To get from %s to %s takes %d knight moves.\n",s,e,GetRe(start,End));      }      return 0;  }