poj A Knight's Journey（DFS）（字典序）

A Knight's Journey

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

ps：
弄了一下午，坑死在了字典序上0.0


这一题的字典序：就是先按列排序，较小的在前。然后按行排序，也是较小的在前。
int diri[8]={-1,1,-2,2,-2,2,-1,1};

int dirj[8]={-2,-2,-1,-1,1,1,2,2};

也就是（借鉴大神的图）



另外，国际象棋的棋盘：行为数字p，列为字母q

代码：

#include<stdio.h>
#include<string.h>
int n,m,flag;
int vis[30][30];
int to[8][2]= {{-1,-2},{1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}};//方向
struct node
{
int x,y;
} w[30];
int judge(int x,int y)
{
if(x>0&&x<=n&&y>0&&y<=m&&!vis[x][y])
return 1;
return 0;
}
void dfs(int x,int y,int step)
{
w[step].x=x,w[step].y=y;
if(step==n*m)//如果已经遍历了所有的点
{
for(int i=1; i<=step; i++)
printf("%c%d",w[i].y-1+'A',w[i].x);
printf("\n");
flag=1;//标记已经找到答案
}
if(flag)
return ;
for(int i=0; i<8; i++)
{
int xx=x+to[i][0];
int yy=y+to[i][1];
if(judge(xx,yy))
{
vis[xx][yy]=1;
dfs(xx,yy,step+1);
vis[xx][yy]=0;
}
}
if(flag)
return ;
}
int main()
{
int t,cont=0;
scanf("%d",&t);
while(++cont<=t)
{
memset(vis,0,sizeof(vis));
flag=0;
cf36
scanf("%d%d",&n,&m);
printf("Scenario #%d:\n",cont);
if(n>8||m>8)
printf("impossible\n");
else
{
vis[1][1]=1;
dfs(1,1,1);
if(!flag)
printf("impossible\n");
}
printf("\n");
}
return 0;
}