Knight Moves

杭电1372

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.

#include<stdio.h>
#include<queue>
using namespace std;
int a[9][9];
int x,y,m,n;
int c[8][2]={2,1, 2,-1, -2,1, -2,-1, 1,2, -1,2, 1,-2, -1,-2};
struct node
{
    int x,y,step;
};
    
int dfs()
{
    int k;
    queue<node>q;
    node cur,next;
    cur.x=x;
    cur.y=y;
    cur.step=0;
    q.push(cur);
    while(!q.empty())
    {
        cur=q.front();
        q.pop();
    
         
        for(k=0;k<8;k++)
        {
            next.x=cur.x+c[k][0];
            next.y=cur.y+c[k][1];
                if(next.x==m&&next.y==n)
                    return cur.step+1;
            if(next.x<=0||next.y<=0||next.x>8||next.y>8||a[next.x][next.y]==1)
                continue;
                 
                next.step=cur.step+1;
                q.push(next);
                a[next.x][next.y]=1;
        
        }
    }
}


int main()
{
    int b1,b2;
    char s1,s2;
    int t;
    while(scanf("%c",&s1)!=-1)
    {
        memset(a,0,sizeof(a));
        scanf("%d ",&b1);
        scanf("%c%d",&s2,&b2);
        getchar();
        x=s1-'a'+1;
        y=b1;
        m=s2-'a'+1;
        n=b2;
        a[x][y]=1;
        if(x==m&&y==n)
        {
            printf("To get from %c%d to %c%d takes %d knight moves.\n",s1,b1,s2,b2,0);
            continue;
        }
        else
            t=dfs();
    printf("To get from %c%d to %c%d takes %d knight moves.\n",s1,b1,s2,b2,t);
    }
    return 0;
}