hdu 1372 Knight Moves (BFS)

Knight Moves

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5296    Accepted Submission(s): 3234

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.  


University of Ulm Local Contest 1996  


Recommend Eddy   |   We have carefully selected several similar problems for you:  

 1 //46MS    288K    1018 B    C++
 2 /*
 4     题意:
 5         在8*8的棋盘上问在初始位置最少移动多少步能到指定位置
 7     BFS:
 8         基础的BFS.移动像象棋的马一样,每一步最多有八个选择 
10 */
11 #include<iostream>
12 #include<queue>
13 using namespace std;
14 struct node{
15     int x,y,cnt;
16 };
17 int g[10][10];
18 int vis[10][10];
19 int mov[8][2]={1,2,2,1,-1,2,2,-1,1,-2,-2,1,-1,-2,-2,-1};
20 int sx,sy,ex,ey;
21 int bfs()
22 {
23     queue<node>Q;
24     memset(vis,0,sizeof(vis));
25     node t={sx,sy,0};
26     vis[sx][sy]=1;
27     Q.push(t);    
28     while(!Q.empty()){
29         t=Q.front();
30         Q.pop();
31         if(t.x==ex && t.y==ey) return t.cnt; 
32         for(int i=0;i<8;i++){
33             node tt=t;
34             tt.cnt++;
35             tt.x+=mov[i][0];
36             tt.y+=mov[i][1];
37             if(tt.x>0&&tt.x<=8 && tt.y>0&&tt.y<=8 && !vis[tt.x][tt.y]){
38                 Q.push(tt);
39                 vis[tt.x][tt.y]=1;
40             }
41         }
42     }
43 }
44 int main(void)
45 {
46     char s[5],e[5];
47     while(cin>>s>>e)
48     {
49         sx=s[0]-'a'+1;
50         sy=s[1]-'0';
51         ex=e[0]-'a'+1;
52         ey=e[1]-'0';
53         printf("To get from %s to %s takes %d knight moves.\n",s,e,bfs());
54     }
55     return 0;
56 }


    原文地址: https://www.cnblogs.com/GO-NO-1/p/3467235.html