POJ2488 A Knight's Journey

Description

Background
The knight is getting bored of seeing the same black and white squares again and again and has decided to make a journey
around the world. Whenever a knight moves, it is two squares in one direction and one square perpendicular to this. The world of a knight is the chessboard he is living on. Our knight lives on a chessboard that has a smaller area than a regular 8 * 8 board, but it is still rectangular. Can you help this adventurous knight to make travel plans?

Problem
Find a path such that the knight visits every square once. The knight can start and end on any square of the board.

Input

The input begins with a positive integer n in the first line. The following lines contain n test cases. Each test case consists of a single line with two positive integers p and q, such that 1 <= p * q <= 26. This represents a p * q chessboard, where p describes how many different square numbers 1, . . . , p exist, q describes how many different square letters exist. These are the first q letters of the Latin alphabet: A, . . .

Output

The output for every scenario begins with a line containing "Scenario #i:", where i is the number of the scenario starting at 1. Then print a single line containing the lexicographically first path that visits all squares of the chessboard with knight moves followed by an empty line. The path should be given on a single line by concatenating the names of the visited squares. Each square name consists of a capital letter followed by a number.
If no such path exist, you should output impossible on a single line.

Sample Input

Sample Output

Scenario #1:
A1

Scenario #2:
impossible

Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4

Source

TUD Programming Contest 2005, Darmstadt, Germany

要求字典序最小，那么暴力枚举起点然后DFS就行。沿途存下路径，最后转化成字符输出。

 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:
",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("
");
55         if(cas<T)printf("
");
56     }
57     return 0;
58 }