A Knight's Journey POJ

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
DFS 字典序

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int maxn = 30;
bool a[maxn][maxn], flag;
int c[] = { -1, 1, -2, 2, -2, 2, -1, 1 },
        d[] = { -2, -2, -1, -1, 1, 1, 2, 2 };//注意顺序
int n, m, end;
char b[60];

bool ok(int x, int y) {
    return x >= 0 && x < n && y >= 0 && y < m;
}

void dfs(int x, int y, int TM, char *s) {
    if (flag)
        return;
    if (TM == end) {
        puts(s);
        flag = true;
        return;
    }
    for (int i = 0; i < 8; i++) {
        int xx = x + c[i], yy = y + d[i];
        if (ok(xx, yy) && !a[xx][yy]) {
            a[xx][yy] = true;
            b[TM] = yy + 'A';
            b[TM+1] = xx + '1';
            dfs(xx, yy, TM+2, s);
            a[xx][yy] = false;
        }
    }
}

int main() {
    int t;
    int cas=0;
    scanf("%d", &t);
    while (t--) {
        printf("Scenario #%d:
",++cas);
        scanf("%d%d", &n, &m);
        end = n * m * 2;
        flag = false;
        for (int j = 0; j < m; j++) {
            for (int i = 0; i < n; i++) {
                memset(a, 0, sizeof(a));
                memset(b, 0, sizeof(b));
                b[0] = j + 'A';
                b[1] = i + '1';
                a[i][j]=true;
                dfs(i, j, 2, b);
                if (flag)
                    goto goubitimulangfeiwoyixiawushijain;
            }

        }
        goubitimulangfeiwoyixiawushijain: if (!flag)
            puts("impossible");
        puts("");
    }
}