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

3
1 1
2 3
4 3

Sample Output

Scenario #1:
A1

Scenario #2:
impossible

Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4

解题思路

水DFS。注意搜索的顺序，因为要求字典序输出！

代码

#include 
   
     #include 
    
      #include 
     
       using namespace std; const int maxn = 30; int vis[maxn][maxn]; int kase = 0; int width,high; int cnt; struct node { int x; char y; }; node s[maxn*maxn]; bool dfs(int x,int y) { if(x < 1 || x > high || y < 1 || y > width) return false; if(vis[x][y] == 1) return false; s[cnt].x = x; vis[x][y] = 1; s[cnt++].y = y+'A'-1; if(cnt == width*high) return true; if(dfs(x-1,y-2) || dfs(x+1,y-2) || dfs(x-2,y-1) || dfs(x+2,y-1) || dfs(x-2,y+1) || dfs(x+2,y+1) || dfs(x-1,y+2) || dfs(x+1,y+2)) return true; cnt --; vis[x][y] = 0; return false; } int main() { int n; scanf("%d",&n); while(n--) { for(int i = 0 ; i < maxn*maxn ; i ++) { s[i].x = 0; s[i].y = 0; } memset(vis,0,sizeof(vis)); scanf("%d%d",&high,&width); printf("Scenario #%d:\n",++kase); cnt = 0; if(!dfs(1,1)) {printf("impossible\n");if(n!=0)printf("\n");continue;} for(int i = 0 ; i < cnt ; i ++) printf("%c%d",s[i].y,s[i].x); printf("\n"); if(n != 0) printf("\n"); } return 0; }