439 - Knight Moves

A friend of you isdoing research on the Traveling Knight Problem (TKP) where youare to find the shortest closed tour of knight moves that visits each square ofa given set of n squares on a chessboard exactly once. Hethinks that the most difficult part of the problem is determining the smallestnumber of knight moves between two given squares and that, once you haveaccomplished this, finding the tour would be easy.

Of course you knowthat it is vice versa. So you offer him to write a program that solves the"difficult" part.

Your job is towrite a program that takes two squares a and b asinput and then determines the number of knight moves on a shortest routefrom a to b.

Input Specification

The input filewill contain one or more test cases. Each test case consists of one linecontaining two squares separated by one space. A square is a string consistingof a letter (a-h) representing the column and a digit (1-8) representing therow on the chessboard.

Output Specification

For each testcase, print one line saying "To get from xx to yy takes n knightmoves.".

Sample Input

e2 e4

a1 b2

b2 c3

a1 h8

a1 h7

h8 a1

b1 c3

f6 f6

Sample Output

To getfrom e2 to e4 takes 2 knight moves.

To get from a1 tob2 takes 4 knight moves.

To get from b2 toc3 takes 2 knight moves.

To get from a1 toh8 takes 6 knight moves.

To get from a1 toh7 takes 5 knight moves.

To get from h8 toa1 takes 6 knight moves.

To get from b1 toc3 takes 1 knight moves.

To get from f6 tof6 takes 0 knight moves.

代码:

#include<iostream>

#include<cstring>

#include<queue>

using namespacestd;

 

struct dot

{

    int r,c;

    dot(int r=0,int c=0):r(r),c(c) {}

};

 

int vis[8][8];

int d[8][8];

int dr[] ={-2,-2,-1,-1, 1,1, 2,2};

int dc[] = {-1,1,-2, 2,-2,2,-1,1};

 

int bfs(int r1,intc1,int r2,int c2);

 

int main()

{

    string s1,s2;

    while(cin>>s1>>s2)

    {

        memset(vis,0,sizeof(vis));

        intans=bfs(s1[0]-'a',s1[1]-'1',s2[0]-'a',s2[1]-'1');

        cout<<"To get from"<<s1<<" to "<<s2<<" takes"<<ans<<" knight moves.\n";

    }

    return 0;

}

 

int bfs(int r1,intc1,int r2,int c2)

{

    if(r1==r2&&c1==c2)

    {

        return 0;

    }

    queue<dot> q;

    q.push(dot(r1,c1));

    vis[r1][c1]=1;

    d[r1][c1]=0;

    while(!q.empty())

    {

        dot a=q.front();

        q.pop();

        for(int i=0;i<8;i++)

        {

            int newr=a.r+dr[i];

            int newc=a.c+dc[i];

           if(newr>=0&&newr<8&&newc>=0&&newc<8&&vis[newr][newc]==0)

            {

                q.push(dot(newr,newc));

                d[newr][newc]=d[a.r][a.c]+1;

                vis[newr][newc]=1;

                if(newr==r2&&newc==c2)

                {

                    return d[newr][newc];

                }

            }

        }

    }

    return -1;

}