poj 2488 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

我觉得这个题的坑爹之处就在于它的顺序问题。。八个方向，从最左边一列开始，从左往右走。至于为什么非要按这个顺序，有待商榷。。。。

 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.。。。波浪线的那个 就是 字典序 的意思。。翻译：然后打印出一行，包含骑士以字典序走遍棋盘所有方格的路径，并且紧跟着一个空行。。。

#include 

int p, q, a[30][30], path[60], flag, l;     //error：注意path的大小，题目给出了p*q<=26

void dfs(int l, int m, int n)

{

    if(l == 2*p*q)

    {

        flag = 1;

        return;

    }

    if(a[m][n] == 1 || m < 2 || n < 2 || m > p+1 || n > q+1)

          return ;

    if(!flag)

    {

       a[m][n] = 1;

    path[l] = n;                   

    path[l+1] = m;

    dfs(l+2, m-1, n-2);          //error：这里两个数都存入一个一维数组的话，下一层的l应直接+2，而非+1

    dfs(l+2, m+1, n-2);

    dfs(l+2, m-2, n-1);

    dfs(l+2, m+2, n-1);

    dfs(l+2, m-2, n+1);

    dfs(l+2, m+2, n+1);

    dfs(l+2, m-1, n+2);

    dfs(l+2, m+1, n+2);

    a[m][n] = 0;

    }

}

int main()

{

    int  t, i, j, x;

    scanf("%d", &t);

    for(x = 1 ; x <= t ; x++)                //error 1：没有输入t就直接输入p，q了，纠结了好一会为毛输入和输出不一样。呵呵。。。

    {

        flag = 0;

        scanf("%d %d", &p, &q);

        printf("Scenario #%d:\n", x);

        for(i = 0 ; i <= p+3 ; i++)

            for(j = 0 ; j <= q+3 ; j++)

            a[i][j] = 0;

        dfs(0, 2, 2);

        if(flag)

        {

            for(i = 0 ; i < 2*p*q ; i++)

           {

           if(i%2 == 0)

                printf("%c", path[i]+63);

           else

                printf("%d", path[i]-1);

            }

            printf("\n");

        }

        else

            printf("impossible\n");

        printf("\n");

    }

    return 0;

}