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.
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.
思路：
先把字母转换成数字储存 再BFS
参考代码：
1 #include <stdio.h> 2 #include <string.h> 3 #include <queue> 4 using namespace std; 5 int movex[8][2]={{-2,1},{-1,2},{1,2},{2,1},{2,-1},{1,-2},{-1,-2},{-2,-1}}; 6 int mark[10][10]; 7 int step[10][10]; 8 int bfs(int a,int b,int c,int d); 9 int main() 10 { 11 int a0,b0,an,bn; 12 char temp1,temp2; 13 while(scanf("%c%d %c%d",&temp1,&b0,&temp2,&bn)!=EOF) 14 { 15 getchar(); 16 memset(mark,0,sizeof(mark)); 17 memset(step,0,sizeof(step)); 18 a0=temp1-‘a‘+1; 19 an=temp2-‘a‘+1; 20 if(temp1==temp2&&b0==bn) 21 { 22 printf("To get from %c%d to %c%d takes 0 knight moves.\n",temp1,b0,temp2,bn); 23 continue; 24 } 25 int ans=bfs(a0,b0,an,bn); 26 printf("To get from %c%d to %c%d takes %d knight moves.\n",temp1,b0,temp2,bn,ans); 27 } 28 return 0; 29 } 30 31 int bfs(int a,int b,int c,int d) 32 { 33 queue<int> qx; 34 queue<int> qy; 35 qx.push(a); 36 qy.push(b); 37 mark[a][b]=1; 38 while(!qx.empty()&&!qy.empty()) 39 { 40 int i; 41 for(i=0;i<8;i++) 42 { 43 int nx=qx.front()+movex[i][0]; 44 int ny=qy.front()+movex[i][1]; 45 if(nx>=1&&nx<=8&&ny>=1&&ny<=8&&!mark[nx][ny]) 46 { 47 mark[nx][ny]=1; 48 step[nx][ny]=step[qx.front()][qy.front()]+1; 49 qx.push(nx); 50 qy.push(ny); 51 if(nx==c&&ny==d) 52 return step[nx][ny]; 53 } 54 } 55 qx.pop(); 56 qy.pop(); 57 } 58 }