Knight Moves

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14125    Accepted Submission(s): 8269

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   |   We have carefully selected several similar problems for you:   1072  1240  1312  1241  1016    -------------------------------------------------------------------------------------------------------------------------------------------------------------------- 本題為典型的bfs深搜的模板題 本題只要瞭解bfs演算法就能AC 但是要對國際象棋中的馬的運動方式熟悉一下就沒問題了 下面上我註釋了的 誰都能看懂的程式碼 --------------------------------------------------------------------------------------------------------------------------------------------------------------------  

  1 //Author:LanceYu
  2 #include<iostream>
  3 #include<string>
  4 #include<cstring>
  5 #include<cstdio>
  6 #include<fstream>
  7 #include<iosfwd>
  8 #include<sstream>
  9 #include<fstream>
 10 #include<cwchar>
 11 #include<iomanip>
 12 #include<ostream>
 13 #include<vector>
 14 #include<cstdlib>
 15 #include<queue>
 16 #include<set>
 17 #include<ctime>
 18 #include<algorithm>
 19 #include<complex>
 20 #include<cmath>
 21 #include<valarray>
 22 #include<bitset>
 23 #include<iterator>
 24 #define ll long long
 25 using namespace std;
 26 const double clf=1e-8;
 27 //const double e=2.718281828;
 28 const double PI=3.141592653589793;
 29 const int MMAX=2147483647;
 30 //priority_queue<int>p;
 31 //priority_queue<int,vector<int>,greater<int> >pq;
 32 struct node
 33 {
 34     int x,y,step;
 35 };
 36 queue<node> q;
 37 int dir[8][2]={{-2,-1},{-1,-2},{-2,1},{-1,2},{1,2},{1,-2},{2,-1},{2,1}};//馬所能夠跳的八個方向記錄下來 
 38 int vis[8][8];
 39 char temp[3][3];//定義一個字串用於輸入 
 40 int change(char c)//字元轉數字 
 41 {
 42     switch (c) 
 43     {
 44         case 'a':return 0;
 45             break;
 46         case 'b':return 1;
 47             break;
 48         case 'c':return 2;
 49             break;
 50         case 'd':return 3;
 51             break;
 52         case 'e':return 4;
 53             break;
 54         case 'f':return 5;
 55             break;
 56         case 'g':return 6;
 57             break;
 58         case 'h':return 7;
 59             break;
 60         case '1':return 0;
 61             break;
 62         case '2':return 1;
 63             break;
 64         case '3':return 2;
 65             break;
 66         case '4':return 3;
 67             break;
 68         case '5':return 4;
 69             break;
 70         case '6':return 5;
 71             break;
 72         case '7':return 6;
 73             break;
 74         case '8':return 7;
 75             break;
 76     }
 77 }
 78 int bfs(int x,int y,int x1,int y1)
 79 {
 80     while(!q.empty())//佇列的初始化，全部清空 
 81         q.pop();
 82     int i;
 83     q.push(node{x,y,0});
 84     while(!q.empty())
 85     {
 86         node t=q.front();
 87         q.pop();
 88         if(t.x==x1&&t.y==y1)
 89             return t.step;
 90         for(i=0;i<8;i++)
 91         {
 92             int dx=t.x+dir[i][0];
 93             int dy=t.y+dir[i][1];
 94             if(dx>=0&&dy>=0&&dx<8&&dy<8&&!vis[dx][dy])//基本搜尋 
 95             {
 96                 vis[dx][dy]=1;
 97                 q.push(node{dx,dy,t.step+1});
 98             }
 99         }
100     }
101     return 0;
102 }
103 int main()
104 {
105     while(scanf("%s%s",temp[0],temp[1])!=EOF)
106     {
107         memset(vis,0,sizeof(vis));
108         int x=change(temp[0][0]);
109         int y=change(temp[0][1]);
110         int x1=change(temp[1][0]);
111         int y1=change(temp[1][1]);//確定首尾點 
112         int ans=bfs(x,y,x1,y1);
113         printf("To get from %s to %s takes %d knight moves.\n",temp[0],temp[1],ans);//輸出 
114     }
115     return 0;
116 }

2018-11-16　　00:03:31　　Author:LanceYu