hdu 1372 骑士从起点走到终点的步数 (BFS)

给出起点和终点 求骑士从起点走到终点所需要的步数

Sample Input
e2 e4 //起点 终点
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 <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
using namespace std;


int map[15][15];
int sx, sy;
int ex, ey;
bool flag;
char s1[5] , s2[5] ;

int dx[] = {-2,-1,1,2,-2,-1,1,2} ;
int dy[] = {1,2,2,1,-1,-2,-2,-1} ;

struct node
{
    int x , y , step ;
};

int bfs()
{
    queue<node> q ;
    int i , fx ,fy ;
    node now , t ;
    now.x = sx ;
    now.y = sy ;
    now.step = 0 ;
    q.push(now) ;
    memset(map , 0 , sizeof(map)) ;
    map[sx][sy] = 1 ;
    while(!q.empty())
    {
        now = q.front() ;
        q.pop() ;
        for (i = 0 ; i < 8 ; i++)
        {
            fx = now.x + dx[i] ;
            fy = now.y + dy[i] ;
            if (fx<0 || fy<0 || fx>= 8 || fy>= 8 || map[fx][fy] == 1)
                continue ;
            if (fx == ex && fy == ey)
            {
                return now.step+1 ;
            }
            t.x = fx ;
            t.y = fy ;
            t.step = now.step+1 ;
            q.push(t) ;
            map[fx][fy] = 1 ;
        }
    }
    return 0 ;
}



int main()
{
    //freopen("in.txt","r",stdin) ;
    while (scanf("%s %s" , &s1 , &s2) !=EOF)
    {
        sx = s1[0] - 'a' ;
        sy = s1[1] - '1' ;
        ex = s2[0] - 'a' ;
        ey = s2[1] - '1' ;
        printf("To get from %s to %s takes %d knight moves.
",s1,s2,bfs()) ;
    }



    return 0 ;
}