题意很简单：
问一个马从起点走到终点最短步数。
思路：
简单的bfs，输入得到起点终点，直接用队列走就可以了！
#include<cstdio>
#include<queue>
using namespace std;
int sx,sy,gx,gy;
const int INF = 1e8;
int idx[10][10];
const int dx[] = {-1,-2,-2,-1,1,2,2,1};
const int dy[] = {-2,-1,1,2,2,1,-1,-2};
typedef pair<int,int>P;
int main()
{
char s1[5],s2[5];
while(scanf("%s%s",s1,s2) == 2){
for (int i = 1 ; i <= 8; ++i)
for (int j = 1; j <= 8; ++j)
idx[i][j]=INF;
queue<P>q;
sy=s1[1]-48;sx=8-(s1[0]-'a');
gy=s2[1]-48;gx=8-(s2[0]-'a');
q.push(P(sx,sy));
idx[sx][sy]=0;
while(!q.empty()){
P p = q.front();q.pop();
if (p.first == gx && p.second == gy)break;
for (int i = 0; i < 8; ++i){
int nx = p.first + dx[i];
int ny = p.second + dy[i];
if (nx >= 1 && nx <= 8 && ny >= 1 && ny <= 8 && idx[nx][ny] == INF){
q.push(P(nx,ny));
idx[nx][ny] = idx[p.first][p.second] + 1;
}
}
}
printf("To get from %s to %s takes %d knight moves.\n",s1,s2,idx[gx][gy]);
}
return 0;
}