hdu 1372 Knight Moves(经典BFS)

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

大意：至少要用多少步，骑士才能从给定位置到达另一个位置。（骑士的走法和象棋里的马的走法是一样的，均是日字型）

Sample Input

Sample Output

e2 e4
a1 b2
b2 c3
a1 h8
a1 h7
h8 a1
b1 c3
f6 f6

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.

显然，这需要BFS。

BFS核心是一次性先把所有最近的能走的统统遍历，然后再去遍历其他的点。这和队列联系在一起了，近点先进，远点后进，对队列的遍历就相当于层次遍历了。比如这个二叉树的层次遍历：

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
bool vis[9][9];
int s[2],e[2];
int dir[8][2]={{1,2},{2,1},{2,-1},{1,-2},{-1,-2},{-2,-1},{-1,2},{-2,1}};
typedef struct node{
    int x,y,step;  //pre[]
}Node;
Node que[90];
int bfs(int sx,int sy){
    int l=0,r=0;
    vis[sx][sy]=1;
    que[r].x=sx;
    que[r].y=sy;
    que[r++].step=0;
    while(l<r){
        Node cur=que[l++],next;
        for(int i=0;i<8;i++){
            next.x=cur.x+dir[i][0];
            next.y=cur.y+dir[i][1];
            next.step=cur.step+1;
            if(next.x>=1&&next.x<=8&&next.y<=8&&next.y>=1&&!vis[next.x][next.y]){
                 if(next.x==e[0]&&next.y==e[1]){  return next.step; }
                 vis[next.x][next.y]=1;
                 que[r++]=next;
            }
        }
    }
    return 0;
}
int main()
{
    //freopen("cin.txt","r",stdin);
    char start[5],end[5];
    while(cin>>start>>end){
         s[0]=start[0]-'a'+1; //加1啊，不要犯错。
         s[1]=start[1]-'0';
         e[0]=end[0]-'a'+1;
         e[1]=end[1]-'0';
         memset(vis,0,sizeof(vis));
         memset(que,0,sizeof(que));
         printf("To get from %s to %s takes %d knight moves.\n",start,end,bfs(s[0],s[1]));
    }
    return 0;
}