[恢]hdu 1372

2011-12-25 09:14:55

地址：http://acm.hdu.edu.cn/showproblem.php?pid=1372

题意：8*8的棋盘，给两个点，问Knight最少走几步能从一个点走到另一个点。

mark：直接bfs即可。

代码：

# include <stdio.h>
# include <string.h>


# define SIZE 8


int graph[70][70] ;
int visited[10][10] ;
int q[1000] ;


int min(int a, int b){return a<b?a:b;}
int abs(int n){return n<0?-n:n;}


void bfs (int idx)
{
    int i, pos ;
    int front = 0, rear = 0 ;
    int x, y, xx, yy, step ;
    int tab[8][2] = {-2, -1, -2, 1, -1, -2, -1, 2, 
                    1, -2, 1, 2, 2, -1, 2, 1} ;
    graph[idx][idx] = 0 ;
    q[rear++] = idx ;
    while (front != rear)
    {
        pos = q[front++] ;
        step = graph[idx][pos] ;
        x = pos/8, y = pos % 8 ;
        for (i = 0 ; i < 8 ; i++)
        {
            xx = x + tab[i][0] ;
            yy = y + tab[i][1] ;
            if (xx < 0 || xx >= 8 || yy < 0 || yy >= 8)
                continue ;
            if (graph[idx][xx*8+yy] < 100) continue ;
            graph[idx][xx*8+yy] = step + 1 ;
            q[rear++] = xx*8+yy ;
        }
    }
}


int main ()
{
    char a,b,c,d ;
    int i, x1, y1, x2, y2 ;
    
    memset (graph, 0x7f, sizeof(graph)) ;
    while (~scanf ("%c%c %c%c%*c", &a, &b, &c, &d))
    {
        x1 = a-'a', y1 = b-'1' ;
        x2 = c-'a', y2 = d-'1' ;
        bfs (x1*8+y1) ;
        printf ("To get from %c%c to %c%c takes %d knight moves.\n",
            a, b, c, d, graph[x1*8+y1][x2*8+y2]) ;
    }
    return 0 ;
}