題目連結：http://acm.hdu.edu.cn/showproblem.php?pid=1372

Knight Moves

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14208    Accepted Submission(s): 8306

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.   題意：在一個 8 × 8 的網格中，給你兩個點，按馬行日的方法走，求走的最短步數，注意下輸出的格式，別隻輸出了結果沒有輸出那句話。。。

注意：馬走日，有八個方向！

 1 #include<iostream>
 2 #include<queue>
 3 #include<cstdio>
 4 #include <cstring>
 5 
 6 using namespace std;
 7 
 8 int go[8][2] = {1,2,2,1,2,-1,1,-2,-1,-2,-2,-1,-2,1,-1,2};
 9 char s1[3],s2[3];
10 int sx,sy,ex,ey;
11 bool mark[10][10];
12 struct node
13 {
14     int
 
 x,y;
15     int step;
16 };
17 
18 void bfs()
19 {
20     queue<node> q;
21     struct node cu,ne;
22     cu.x = sx;
23     cu.y = sy;
24     cu.step =0;
25     mark[cu.x][cu.y] = 1; //別忘了標記，雖然本題不標記也可以；
26     q.push(cu);
27     while(!q.empty())
28     {
29         cu = q.front();
30         q.pop();
 
31         if(cu.x == ex && cu.y == ey)
32         {
33             printf("To get from %s to %s takes %d knight moves.\n",s1,s2,cu.step);
34             return ;
35         }
36         for(int i=0;i<8;i++)
37         {
38             ne.x = cu.x + go[i][0];
39             ne.y = cu.y + go[i][1];
40             if(ne.x>=0&&ne.x<8&&ne.y>=0&&ne.y<8&&!mark[ne.x][ne.y]) //注意這個範圍是0--7的；
41             {
42                 ne.step = cu.step + 1;
43                 mark[ne.x][ne.y] = 1;
44                 q.push(ne);
45             }
46         }
47     }
48 }
49 int main()
50 {
51     //也可以用字母加數字的方式輸入。eg while(scanf("%c%d %c%d",&c1,&d1,&c2,&d2),但別忘了輸出也是字母加數字
52     while(scanf("%s%s",s1,s2)!=EOF) //這兩個%s之間的空格有沒有都可以
53     {
54         //a--h 和 1--8 數量是一樣的，所以，a--h減'a'相當於0--7，所以，數字別忘了減 1 ；
55         sx = s1[0] - 'a';
56         sy = s1[1] - '1';
57         ex = s2[0] - 'a';
58         ey = s2[1] - '1';
59         memset(mark,0,sizeof(mark));
60         bfs();
61         memset(s1,0,sizeof(s1));
62         memset(s2,0,sizeof(s2));
63     }
64     return 0;
65 }

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