poj2488

                                                                                  A Knight's Journey

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 25463		Accepted: 8672

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

#include<iostream>
#include<cstring>
#include<cstdio>
#include<string>
#include<algorithm>
using namespace std;
int dir[8][2]={-1,-2, 1,-2, -2,-1, 2,-1, -2,1, 2,1, -1,2, 1,2};
bool flag ;
int p,q,i;    int n;
const int maxn = 30;
int sign[30][30];
struct node 
{
    int x,y;
}path[maxn];
void dfs(int x,int y,int step)
{
    sign[x][y] = 1;
    path[step].x = x;path[step].y = y;
    if(step==p*q-1)
    {
        flag = 1;
        for(i=0;i<=step;i++)
        {
            printf("%c%d",path[i].y+'A',path[i].x+1);
        }
        printf("\n\n");
    }
//    printf("%c%d\n",path[step].y+'A',path[step].x+1);
    for(i=0;i<8;i++)
    {
        int xx = x + dir[i][0];
        int yy = y + dir[i][1];
        if(xx>=p||yy>=q||xx<0||yy<0)
            continue; 
        if(!sign[xx][yy])
        {
            sign[xx][yy] = 1;
            dfs(xx,yy,step+1);
            sign[xx][yy] = 0;
        }
    }
}
int main()
{
    
    scanf("%d",&n);
    int h =1;
    while(n--)
    {
        flag = 0;
        scanf("%d %d",&p,&q);
        memset(sign,0,sizeof(sign));
        printf("Scenario #%d:\n",h++);
        dfs(0,0,0);
        if(!flag)
        {
            printf("impossible\n\n");
        }
    }
    return 0;
}