题目地址：​​点击打开链接​​

思路：简单BFS，可以走8个方向

AC代码：

#include<iostream>
#include<queue>
#include<cstdio>
using namespace std;
struct point
{
  int x,y;
  int c;
}from,to;
int dx[8]={-1,-2,-2,-1,1,2,2,1};
int dy[8]={-2,-1,1,2,-2,-1,1,2};
void bfs(char a[],char b[])
{
  point temp;
  int i;
  queue<point> q;
  from.x=a[0]-'a';
  from.y=a[1]-'1';
  from.c=0;
  to.x=b[0]-'a';
  to.y=b[1]-'1';
  q.push(from);
  while(true)
  {
    from=q.front();
    q.pop();
    if(from.x==to.x&&from.y==to.y)
      break;
    for(i=0;i<8;i++)
    {
      temp.x=from.x+dx[i];
      temp.y=from.y+dy[i];
      temp.c=from.c+1;
      if(temp.x<0||temp.x>7||temp.y<0||temp.y>7)
        continue;
      q.push(temp);
    }
  }
}
int main()
{
  char a[3],b[3];
  while(scanf("%s%s",a,b)!=EOF)
  {
    bfs(a,b);
    printf("To get from %s to %s takes %d knight moves.\n",a,b,from.c);
  }
  return 0;
}