Zju1091 Knight Moves

1284: Zju1091 Knight Moves

Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 86  Solved: 54
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Description

在一个8*8的棋盘上，一只中国象棋中的马要从一个点跳到另一个点。问最少需要多少步。

Input

整个测试组由多组数据组成，请做到文件底结束。
对于每组数据，前两个坐标代表出发点，后两个代表结束点。注意坐标第一位为a至h中某个字母，第二位为1到8某个数字。

Output

对于每个测试请输出"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.

HINT

 

Source

#include <iostream>
#include <cstring>
using namespace std;
int u[8]={-2,-2,-1,-1,1,1,2,2};
int v[8]={-1,1,-2,2,-2,2,-1,1};
int sx,sy,ex,ey;
int g[9][9],x[90],y[90];

void bfs()
{
     int t,w,tx,ty,i;
     memset(g,0,sizeof(g));
     x[1]=sx; 
     y[1]=sy;
     g[sx][sy]=1;
     t=0;
     w=1;
     do
     {
           t++;
           for (i=0;i<8;i++)
           {
               tx=x[t]+u[i]; 
               ty=y[t]+v[i];
               if (tx<1 || tx>8 || ty<1 || ty>8 || g[tx][ty]!=0) 
               continue;
               g[tx][ty]=g[x[t]][y[t]]+1;
               w++;
               x[w]=tx; 
               y[w]=ty;
               if (tx==ex && ty==ey) 
               return;
           }
     } while (t!=w);
}

int main()
{
    char c[4];
    int i,ans;
    while (cin>>c[0]>>c[1]>>c[2]>>c[3])
    {
          sx=c[0]-'a'+1; 
          sy=c[1]-'0';
          ex=c[2]-'a'+1; 
          ey=c[3]-'0';
          if (sx==ex && sy==ey) 
          ans=0;
          else
          {
              bfs(); 
              ans=g[ex][ey]-1;
          }
          cout<<"To get from "<<c[0]<<c[1]<<" to "<<c[2]<<c[3]<<" takes "<<ans<<" knight moves.";
          cout<<endl;
    }
    return 0 ;
}