C

Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

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

经典的DFS，大概题意就是让骑士走遍棋盘的所有格子。一路走到黑，看这条路能不能使骑士把格子都走完。

 1 #include<cstdio>
 2 #include<string.h>
 3 using namespace std;
 4 int vis[30][30];
 5 char str[2000];
 6 int s[8][2]={-2,-1,-2,1,-1,-2,-1,2,1,-2,1,2,2,-1,2,1};//这种处理方式很常用的，要记住，即骑士当前所在位置的周围所能走到的位置
 7 int p,q;
 8 int dfs(int x,int y,int sum,int cnt)
 9 {
10     if(sum==p*q) return 1;//sum是记录走过的格子
11     int x1,y1;
12     for(int i=0;i<8;i++)
13     {
14         x1=x+s[i][0];
15         y1=y+s[i][1];
16         if(x1>=0&&x1<q&&y1>=0&&y1<p&&!vis[x1][y1])//边界条件及判断是否访问过吗
17         {
18             vis[x1][y1]=1;
19             str[cnt+1]=x1+'A';//开一个str数组，用来记录
20             str[cnt+2]=y1+'1';
21             if(dfs(x1,y1,sum+1,cnt+2))//记得这里要写成sum+1,cnt+2
22                 return 1;
23             vis[x1][y1]=0;//回溯
24         }
25     }
26     return 0;
27 }
28 int main()
29 {
30     int t;
31     scanf("%d",&t);
32     for(int i=1;i<=t;i++)
33     {
34         scanf("%d %d",&p,&q);
35         memset(vis,0,sizeof(vis));
36         memset(str,0,sizeof(str));
37         str[0]='A';
38         str[1]='1';
39         vis[0][0]=1;
40         if(dfs(0,0,1,1)){
41             printf("Scenario #%d:
",i);
42             for(int j=0;j<strlen(str);j++)
43                 printf("%c",str[j]);
44             printf("

");//记得每组数据输出后，有个空行
45         }
46         else{
47             printf("Scenario #%d:
",i);
48             printf("impossible

");
49         }
50     }
51     return 0;
52 }