@[LukeSu](/user/593753)
```cpp
#include<bits/stdc++.h>
using namespace std;
const int N = 66;
int g[N][N];
int maxx = 8, maxy = 8; //定义棋盘的边界
char ch1, temp1, ch2, temp2; //获取起点输入，终点输入
int bx, by, ex, ey; //起点位置和终点位置
int dx[8] = {1, 1, -1, -1, 2, 2, -2, -2}; //马的步伐
int dy[8] = {2, -2, 2, -2, -1, 1, 1, -1};
queue< pair<int, int> > q;
int st[N][N] = {};
int dist[N][N]; //记录步数
int xx, yy;
void bfs(){
while(!q.empty()){
int x = q.front().first;
int y = q.front().second;
q.pop();
st[x][y] = 1;
for(int i = 0; i < 8; i++){
xx = dx[i] + x;
yy = dy[i] + y;
if(xx >= 1 and xx <= maxx and yy >= 1 and yy <= maxy and st[xx][yy] == 0){
dist[xx][yy] = dist[x][y] + 1;
if(xx == ex and yy == ey){
return;
}
q.push(make_pair(xx, yy));
}
}
}
}
int main(){
map<char, int> mp;
mp['a'] = 1; mp['b'] = 2; mp['c'] = 3; mp['d'] = 4;
mp['e'] = 5; mp['f'] = 6; mp['g'] = 7; mp['h'] = 8;
while(scanf("%c%d %c%d", &ch1, &temp1, &ch2, &temp2) != EOF){
bx = mp[ch1];
by = temp1;
ex = mp[ch2];
ey = temp2;
dist[ex][ey] = 0;
q.push(make_pair(bx, by));
bfs();
cout << "To get from " << ch1 << by << " to " << ch2 << ey << " takes " << dist[xx][yy] << " knight moves.";
memset(st, 0, sizeof st);
while(!q.empty()) q.pop();
memset(dist, 0, sizeof dist);
}
return 0;
}
```
by LukeSu @ 2022-03-04 11:54:58
@[LukeSu](/user/593753)
```
while(scanf("%c%d %c%d", &ch1, &temp1, &ch2, &temp2) != EOF){
```
可以改成
```
while(scanf("%c%d %c%d\r\n", &ch1, &temp1, &ch2, &temp2) != EOF){
```
然后在cout最后加上\n
by 编码落寞 @ 2022-03-04 12:52:35
@[编码落寞](/user/557751)
按你的方法第一行输入数据读不进去。。
by LukeSu @ 2022-03-04 20:26:56
@[LukeSu](/user/593753)
你在洛谷IDE上试，因为\r\n是linux下换行符
by 编码落寞 @ 2022-03-07 10:49:59
@[编码落寞](/user/557751)
问题我解决了！
使用rewind方法放在while里即可，rewind()作用是清空缓存区，这样就可以避免读入多余换行符：
```cpp
rewind(stdin);
```
by LukeSu @ 2022-03-07 19:39:05
@[编码落寞](/user/557751)
样例是过了，可是提交完TLE了。。。。(笑哭)
看来我还是得换种思路来解决这道题。总之谢谢你的帮助！
by LukeSu @ 2022-03-07 19:41:24
不知道楼主现在解决这个问题没有，过了好久了（
你可以试试把输入改成 ```
scanf("%c%d %c%d%*c", &ch1, &temp1, &ch2, &temp2) != EOF)```
我这么改就好了
by Eureka_yzy @ 2022-06-06 22:35:51