A Knight's Journey (dfs)

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 

题意：给出一个p行q列的国际棋盘，马可以从任意一个格子开始走，问马能否不重复的走完所有的棋盘。如果可以，输出按字典序排列最小的路径。打印路径时，列用大写字母表示（A表示第一列），行用阿拉伯数字表示（从1开始），先输出列，再输出行。

分析：如果马可以不重复的走完所有的棋盘，那么它一定可以走到A1这个格子。所以我们只需从A1这个格子开始搜索，就能保证字典序是小的；除了这个条件，我们还要控制好马每次移动的方向，控制方向时保证字典序最小（即按照下图中格子的序号搜索）。控制好这两个条件，直接从A1开始深搜就行了。

#include<stdio.h>
int n,m;
int vis[105][105];
int lin[105][2];//存放走过的路径
int flag;
int dir[8][2]={{-2,-1},{-2,1},{-1,-2},{-1,2},{1,-2},{1,2},{2,-1},{2,1}};//字典序的方向数组
int dfs(int a,int x,int y)
{

    if(a==m*n)
    {
        flag=1;
        return 1;
    }
    for(int i=0;i<8;i++)
    {
        int xx,yy;
        xx=dir[i][0]+x;
        yy=dir[i][1]+y;

        if(xx>=0 && yy>=0 && xx<m && yy<n && !vis[xx][yy])
        {
            vis[xx][yy]=1;
            lin[a][0]=xx;
            lin[a][1]=yy;
            dfs(a+1,xx,yy);
            if(flag)
                return 1;
            vis[xx][yy]=0;
        }
    }
    return 0;
}

int main()
{
    int t,k=1;
    scanf("%d",&t);
    while(t--)
    {
        for(int i=0;i<105;i++)
            for(int j=0;j<105;j++)
                vis[i][j]=0;
        flag=0;
        scanf("%d%d",&n,&m);
        vis[0][0]=1;

        lin[0][1]=0;
        lin[0][0]=0;



        printf("Scenario #%d:\n",k++);

        if(dfs(1,0,0))
        {
            for(int i=0;i<n*m;i++)
            {
                printf("%c%d",'A'+lin[i][0],lin[i][1]+1);
            }
            printf("\n");
        }

        else
            printf("impossible\n");
        printf("\n");
    }
}