POJ 2488 A Knight's Journey (DFS)

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 
#include 
#define N 30
struct point
{
	int x, y;
};
point map[N][N];
int sum;
int p, q;
int dx[] = { -2, -2, -1, -1, 1, 1, 2, 2 };
int dy[] = { -1, 1, -2, 2, -2, 2, -1, 1 };
bool dfs(int x, int y, int step)
{
	if (step == sum) return true;	
	int newx, newy;
	for (int i = 0; i < 8; i++)
	{
		newx = x + dx[i]; newy = y + dy[i];
		if (newx < 1 || newx > q || newy < 1 || newy > p || map[newx][newy].x != -1) continue;
		map[x][y].x = newx; map[x][y].y = newy;
		if (dfs(newx, newy, step + 1)) return true;
		map[x][y].x = -1; map[x][y].y = -1;
	}
	return false;
}
void route(int x, int y)
{
	if (x == -1) return;
	printf("%c%d", x + 'A' - 1, y);
	route(map[x][y].x, map[x][y].y);
}
int main()
{
	int T;
	scanf("%d", &T);
	for (int I = 1; I <= T; I++)
	{
		scanf("%d%d", &p, &q);
		printf("Scenario #%d:\n", I);
		sum = p * q;
		memset(map, -1, sizeof(map));

		bool flag = false;
		for (int i = 1; i <= q; i++)
		{
			for (int j = 1; j <= p; j++)
			{
				if (dfs(i, j, 1))
				{
					route(i, j);
					flag = true;
					break;
				}
			}
			if (flag) break;
		}
		if (!flag) printf("impossible");
		printf("\n\n");
	}
	return 0;
}