题目大意：在一个8*8的国际想起内，你是国际象棋中马这个角色，然后行从1-8，列从a-h编号，给出起点和终点，问你的最短路径。（马走日子）
思路：简单的bfs，被scanf（“%s%s”）卡了两次，玄学，改成scanf(" %c %c %c %c",&s1[0],&s1[1],&s2[0],&s2[1])就过来，思路还是挺简单的。
代码：

#include<iostream>
#include<queue>
#include<cstring>
using namespace std;

int maze[9][9];
int vis[9][9];
char s1[2],s2[2];
int dir[8][2]={{-1,-2},{-2,-1},{-2,1},{-1,2},{1,2},{2,1},{2,-1},{1,-2}};
struct node{
	int x,y;
	node(int a,int b):x(a),y(b){}
	node(){} 
};
bool inside(int x,int y){
	return (x>=1&&x<=8&&y>=1&&y<=8);
}
int main(){
	int id1,id2;
	while(scanf(" %c %c %c %c",&s1[0],&s1[1],&s2[0],&s2[1])!=EOF){
		memset(maze,0,sizeof(maze));
		memset(vis,0,sizeof(vis));
		sscanf(&s1[1],"%d",&id1);
		sscanf(&s2[1],"%d",&id2);
		queue<node>st;
		st.push(node(id1,s1[0]-'a'+1));
		while(!st.empty()){
			node now=st.front();
			st.pop();
			if(now.x==id2&&now.y==(s2[0]-'a'+1)){
				cout<<"To get from "<< s1[0]<<s1[1]<< " to "<< s2[0]<<s2[1]<<" takes "<< maze[now.x][now.y]<<" knight moves."<<endl;
				break;
			}
			for(int i=0;i<8;i++){
				int fx=now.x+dir[i][0];
				int fy=now.y+dir[i][1];
				if(!vis[fx][fy]&&inside(fx,fy)){
					st.push(node(fx,fy));
					vis[fx][fy]=1;
					maze[fx][fy]=maze[now.x][now.y]+1;
				}
			}
		}
	}
}