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: A1B3C1A2B4C2A3B1C3A4B2C4DFS 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:\n",++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 }
A Knight's Journey POJ - 2488