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.
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:
A1B3C1A2B4C2A3B1C3A4B2C4code:
#includeint m = 0, n = 0, ans[30], used[30][30], step = 0, flag = 0;
int dir[8][2] = {{-1,-2},{1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}};//按字典顺序从（1，1）开始按照这个路径走
void dfs(int x, int y)
{
int i = 0, fx = 0, fy = 0;
if(flag)
return;
if(step == m*n)
{
flag = 1;
return ;
}
else
{
for(i = 0; i<8; i++)
{
fx = x+dir[i][0]; fy = y+dir[i][1];
if(!used[fx][fy] && fx>0 && fx<=m && fy>0 && fy<=n && !flag)
{
used[fx][fy] = 1;
ans[step++] = fx*99+fy;
dfs(fx, fy);
step--;
used[fx][fy] = 0;
}
}
}
}
int main()
{
int i = 1, j = 0, t = 0, count = 0;
scanf("%d",&t);
while(t--)
{
scanf("%d %d",&m,&n);
for(i = 0; i或
#include
int step = 0, count = 0, sign = 0;
int p = 0, q = 0;
int move[26][26];
int dir[8][2] = {{-2, -1}, {-2, 1}, {-1, -2}, {-1, 2},{1, -2}, {1, 2}, {2, -1}, {2, 1}};
typedef struct
{
int x,y;
}node;
node seq[30];
void dfs(int i, int j);
int main()
{
int i = 0, j = 0, n = 0, count = 1;
scanf("%d",&n);
while(n--)
{
step = 0;
sign = 0;
scanf("%d %d",&q,&p);
for(i = 0; i<=p; i++)
for(j = 0; j<=q; j++)
move[i][j] = 0;
dfs(1,1);
if(sign)
{
printf("Scenario #%d:\n",count++);
for(i = 0; i0 && x<=q && y>0 && y<=p)
{
dfs(x,y);
step--;
}
}
move[i][j] = 0;
}