bfs

题目：

A friend of you is doing research on the Traveling Knight Problem (TKP) where you are to find theshortest closed tour of knight moves that visits each square of a given set of n squares on a chessboardexactly once. He thinks that the most difficult part of the problem is determining the smallest numberof knight moves between two given squares and that, once you have accomplished this, finding the tourwould 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 thenumber 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 twosquares separated by one space. A square is a string consisting of a letter (a..h) representing the columnand 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.

题意：从一点走到另一点，a-e代表列，1-8代表行，输入两个坐标，求从第一个坐标走到第二个坐标的步数。走法按罗马象棋走，八个方向，只能走“L”形，即上走两步，左走一步等八个方向。bfs

#include<cstdio>
#include<cstring>
#include<math.h>
#include<algorithm>
#include<queue>
#include<vector>
#include<iostream>
#include<map>
#include<queue>
#define mes(a,b) memset(a,b,sizeof(a))
#define rep(i,m,n) for(i=m;i<=n;i++)
typedef long long ll;
using namespace std;
int max3(int a,int b,int c){return max(max(a,b),c);}
ll min3(ll a,ll b,ll c){return min(min(a,b),c);}
const double PI=acos(-1);
const int inf=0x3f3f3f3f;
const double esp=1e-6;
const int maxn=1e6+5;
const int mod=1e9+7;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll inv(ll b){if(b==1)return 1; return (mod-mod/b)*inv(mod%b)%mod;}
ll fpow(ll n,ll k){ll r=1;for(;k;k>>=1){if(k&1)r=r*n%mod;n=n*n%mod;}return r;}
ll Fpow(ll n,ll k){ll r=1;for(;k;k>>=1){if(k&1)r=r*n;n=n*n;}return r;}

char s1[4],s2[4];
int vis[10][10];
int dir[8][2] = {-2,1,-1,2,1,2,2,1,2,-1,1,-2,-1,-2,-2,-1};
int endx,endy;
struct node
{
    int x,y,step;
};
int bfs()
{
    queue<node>q;
    node p,next,r;
    p.x=s1[0]-'a';
    p.y=s1[1]-'1';
    p.step=0;
    endx=s2[0]-'a';
    endy=s2[1]-'1';
    q.push(p);
    memset(vis,0,sizeof(vis));
    vis[p.x][p.y]=1;
    while(!q.empty())
    {
        r=q.front();
        q.pop();
        if(r.x==endx&&r.y==endy)
        {
            return r.step;
        }
        int i;
        for(i=0;i<8;i++)
        {
            next.x=r.x+dir[i][0];
            next.y=r.y+dir[i][1];
            if(next.x==endx&&next.y==endy)
                return r.step+1;
            if(next.x>=0&&next.x<8&&next.y>=0&&next.y<8&&!vis[next.x][next.y])
            {
                next.step=r.step+1;
                vis[next.x][next.y]=1;
                q.push(next);
            }
        }
    }
    return 0;
}
int main()
{
    while(scanf("%s%s",s1,s2)!=EOF)
    {
        int step=bfs();
        printf("To get from %s to %s takes %d knight moves.\n",s1,s2,step);
    }


    return 0;
}