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<cstdio>
#include<cstring>
#include<string>
#include<set>
#include<vector>
#include<stack>
#include<queue>
#include<algorithm>
#include<cstdio>
#include<algorithm>
#include<functional>
#include<sstream>
int n1, n2;
int xx[30], yy[30];
int visit[30][30];
int flag;
int dir[8][2] = { { -2,-1 },{ -2,1 },{ -1,-2 },{ -1,2 },{ 1,-2 },{ 1,2 },{ 2,-1 },{ 2,1 } };
int go(int x, int y)
{
if (1 <= x&&x <= n1 && 1 <= y&&y <= n2)
return 1;
return 0;
}
void dfs(int x, int y, int num)
{
int i, xxx, yyy;
xx[num] = x;
yy[num] = y;
if (num == n1*n2)
{
flag = 1;
return;
}
for (i = 0; i<8; i++)
{
xxx = x + dir[i][0];
yyy = y + dir[i][1];
if (go(xxx, yyy) && !visit[xxx][yyy] && !flag)
{
visit[xxx][yyy] = 1;
dfs(xxx, yyy, num + 1);
visit[xxx][yyy] = 0;
}
}
}
int main()
{
int i, k, t;
scanf("%d", &t);
for (k = 1; k <= t; k++)
{
scanf("%d%d", &n2, &n1);
memset(visit, 0, sizeof(visit));
visit[1][1] = 1;
flag = 0;
dfs(1, 1, 1);
printf("Scenario #%d:\n", k);
if (flag)
{
for (i = 1; i <= n1*n2; i++)
printf("%c%d", xx[i] + ‘A‘ - 1, yy[i]);
}
else
printf("impossible");
printf(k == t ? "\n" : "\n\n");
}
return 0;
}
A Knight's Journey （DFS）