POJ2488 A Knight's Journey

 1 /**/
 2 #include<iostream>
 3 #include<cstdio>
 4 #include<cmath>
 5 #include<cstring>
 6 #include<algorithm>
 7 using namespace std;
 8 const int mxn=125;
 9 int vis[mxn][mxn];
10 bool flag;
11 int mx[9]={0,-2,-2,-1,-1,1,1,2,2},
12     my[9]={0,-1,1,-2,2,-2,2,-1,1};
13 int n,m;
14 int qx[mxn],qy[mxn];
15 void dfs(int x,int y,int cnt){
16     qx[cnt]=x;qy[cnt]=y;
17     vis[x][y]=1;
18     if(cnt==m*n){
19         flag=1;
20         return;
21     }
22     for(int i=1;i<=8;i++){
23         int nx=mx[i]+x;
24         int ny=my[i]+y;
25         if(nx<1 || nx>m || ny<1 || ny>n)continue;
26         if(vis[nx][ny])continue;    
27         dfs(nx,ny,cnt+1);
28         if(flag)return;
29     }
30     vis[x][y]=0;
31     return;
32 }
33 int main(){
34     int T;
35     scanf("%d",&T);
36     int cas=0;
37     for(cas=1;cas<=T;cas++){
38         memset(vis,0,sizeof vis);
39         flag=0;
40         int i,j;
41         scanf("%d%d",&n,&m);
42         for(i=1;i<=m;i++){
43             if(flag)break;
44              for(j=1;j<=n;j++){
45                  dfs(i,j,1);
46                  if(flag)break;
47              }
48         }
49         printf("Scenario #%d:\n",cas);
50         if(flag){
51             for(i=1;i<=n*m;i++)printf("%c%d",qx[i]+‘A‘-1,qy[i]);
52         }
53         else printf("impossible");
54         printf("\n");
55         if(cas<T)printf("\n");
56     }
57     return 0;
58 }