A Knight's Journey
Time Limit: 1000MS?? Memory Limit: 65536K
Total Submissions: 24580?? Accepted: 8309

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 3Sample Output

Scenario #1:
A1

Scenario #2:
impossible

Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4
Source

TUD Programming Contest 2005, Darmstadt, Germany
[cpp]
#include ??
#include??
int dx[8]= {-1,1,-2,2,-2,2,-1,1};//按字典顺序走必须这么写??
int dy[8]= {-2,-2,-1,-1,1,1,2,2};?
int visit[50][50],way[100][2];//visit记录是否访问。way记录路径??
int p,q;?
?
int dfs(int x,int y,int r)//x,y记录当前所在位置。r记录下一步为第几步??
{?
??? int nx,ny,i;?
?
??? if(r-1==p*q)//如果已经走完则输出??
??????? return 1;?
??? for(i=0; i<8; i++)?
??? {?
??????? nx=x+dx[i];?
??????? ny=y+dy[i];?
??????? if(nx<1||nx>p||ny<1||ny>q||visit[nx][ny])?
??????????? continue;?
??????? visit[nx][ny]=1;?
??????? way[r][0]=nx;?
??????? way[r][1]=ny;?
??????? if(dfs(nx,ny,r+1))?
??????????? return 1;?
??????? visit[nx][ny]=0;?
??? }?
??? return 0;?
}?
int main()?
{?
??? int t,i,k;?
?
??? scanf("%d",&t);?
?
??? for(k=1; k<=t; k++)?
??? {?
??????? scanf("%d%d",&p,&q);?
??????? memset(visit,0,sizeof visit);?
??????? way[1][0]=way[1][1]=1;?
??????? visit[1][1]=1;?
??????? printf("Scenario #%d:\n",k);?
??????? if(dfs(1,1,2))?
??????? {?
??????????? for(i=1; i<=p*q; i++)?
??????????????? printf("%c%d",way[i][1]-1+'A',way[i][0]);?
??????????? printf("\n");?
??????? }?
??????? else?
??????????? printf("impossible\n");?
??????? if(k!=t)?
??????????? printf("\n");?
??? }?
??? return 0;?
}?

#include
#include
int dx[8]= {-1,1,-2,2,-2,2,-1,1};//按字典顺序走必须这么写
int dy[8]= {-2,-2,-1,-1,1,1,2,2};
int visit[50][50],way[100][2];//visit记录是否访问。way记录路径
int p,q;

int dfs(int x,int y,int r)//x,y记录当前所在位置。r记录下一步为第几步
{
??? int nx,ny,i;

??? if(r-1==p*q)//如果已经走完则输出
??????? return 1;
??? for(i=0; i<8; i++)
??? {
??????? nx=x+dx[i];
??????? ny=y+dy[i];
??????? if(nx<1||nx>p||ny<1||ny>q||visit[nx][ny])
??????????? continue;
??????? visit[nx][ny]=1;
??????? way[r][0]=nx;
??????? way[r][1]=ny;
??????? if(dfs(nx,ny,r+1))
??????????? return 1;
??????? visit[nx][ny]=0;
??? }
??? return 0;
}
int main()
{
??? int t,i,k;

??? scanf("%d",&t);