地址：

题意：按照国际象棋规则马最少几步从开始点走到终点。

mark：bfs，仔细。

代码：

#include 
     
      #include 
      
       int a[100][2],d[8][8];char p[5],q[5];void bfs(){    int front = 0,rear = 1,m,n;    d[p[0]-'a'][p[1]-'0'-1] = 0;    a[0][0] = p[0]-'a';    a[0][1] = p[1]-'0'-1;    while(front != rear)    {        m = a[front][0];        n = a[front++][1];        if(m == q[0]-'a' && n == q[1]-'0'-1) return ;        if(m > 1 && n > 0 && !d[m-2][n-1]) a[rear][0] = m-2, a[rear++][1] = n-1, d[m-2][n-1] = d[m][n]+1;        if(m > 0 && n > 1 && !d[m-1][n-2]) a[rear][0] = m-1, a[rear++][1] = n-2, d[m-1][n-2] = d[m][n]+1;        if(m > 1 && n < 7 && !d[m-2][n+1]) a[rear][0] = m-2, a[rear++][1] = n+1, d[m-2][n+1] = d[m][n]+1;        if(m > 0 && n < 6 && !d[m-1][n+2]) a[rear][0] = m-1, a[rear++][1] = n+2, d[m-1][n+2] = d[m][n]+1;        if(m < 7 && n > 1 && !d[m+1][n-2]) a[rear][0] = m+1, a[rear++][1] = n-2, d[m+1][n-2] = d[m][n]+1;        if(m < 6 && n > 0 && !d[m+2][n-1]) a[rear][0] = m+2, a[rear++][1] = n-1, d[m+2][n-1] = d[m][n]+1;        if(m < 7 && n < 6 && !d[m+1][n+2]) a[rear][0] = m+1, a[rear++][1] = n+2, d[m+1][n+2] = d[m][n]+1;        if(m < 6 && n < 7 && !d[m+2][n+1]) a[rear][0] = m+2, a[rear++][1] = n+1, d[m+2][n+1] = d[m][n]+1;    }}int main(){//    freopen("in.txt", "r", stdin);    while(~scanf("%s %s", p, q))    {        memset(d, 0, sizeof(d));        bfs();        printf("To get from %s to %s takes %d knight moves.\n", p, q, d[q[0]-'a'][q[1]-'0'-1]);    }    return 0;}