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

Sample Output

Scenario #1:
A1

Scenario #2:
impossible

Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4
DFS 字典序

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 using namespace std;
 6 
 7 const int maxn = 30;
 8 bool a[maxn][maxn], flag;
 9 int c[] = { -1, 1, -2, 2, -2, 2, -1, 1 },
10         d[] = { -2, -2, -1, -1, 1, 1, 2, 2 };//注意顺序
11 int n, m, end;
12 char b[60];
13 
14 bool ok(int x, int y) {
15     return x >= 0 && x < n && y >= 0 && y < m;
16 }
17 
18 void dfs(int x, int y, int TM, char *s) {
19     if (flag)
20         return;
21     if (TM == end) {
22         puts(s);
23         flag = true;
24         return;
25     }
26     for (int i = 0; i < 8; i++) {
27         int xx = x + c[i], yy = y + d[i];
28         if (ok(xx, yy) && !a[xx][yy]) {
29             a[xx][yy] = true;
30             b[TM] = yy + 'A';
31             b[TM+1] = xx + '1';
32             dfs(xx, yy, TM+2, s);
33             a[xx][yy] = false;
34         }
35     }
36 }
37 
38 int main() {
39     int t;
40     int cas=0;
41     scanf("%d", &t);
42     while (t--) {
43         printf("Scenario #%d:
",++cas);
44         scanf("%d%d", &n, &m);
45         end = n * m * 2;
46         flag = false;
47         for (int j = 0; j < m; j++) {
48             for (int i = 0; i < n; i++) {
49                 memset(a, 0, sizeof(a));
50                 memset(b, 0, sizeof(b));
51                 b[0] = j + 'A';
52                 b[1] = i + '1';
53                 a[i][j]=true;
54                 dfs(i, j, 2, b);
55                 if (flag)
56                     goto goubitimulangfeiwoyixiawushijain;
57             }
58 
59         }
60         goubitimulangfeiwoyixiawushijain: if (!flag)
61             puts("impossible");
62         puts("");
63     }
64 }

View Code