总时间限制: 1000ms 内存限制: 65536kB
描述
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.
输入
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, . . .
输出
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.
样例输入
3
1 1
2 3
4 3
样例输出
Scenario #1:
A1
Scenario #2:
impossible
Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4
来源
TUD Programming Contest 2005, Darmstadt, Germany
大致题意：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。
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 using namespace std; 5 6 int path[88][88], vis[88][88], p, q, cnt; 7 bool flag; 8 9 int dx[8] = {-1, 1, -2, 2, -2, 2, -1, 1}; 10 int dy[8] = {-2, -2, -1, -1, 1, 1, 2, 2}; 11 12 bool judge(int x, int y) 13 { 14 if(x >= 1 && x <= p && y >= 1 && y <= q && !vis[x][y] && !flag) 15 return true; 16 return false; 17 } 18 19 void DFS(int r, int c, int step) 20 { 21 path[step][0] = r; 22 path[step][1] = c; 23 if(step == p * q) 24 { 25 flag = true; 26 return ; 27 } 28 for(int i = 0; i < 8; i++) 29 { 30 int nx = r + dx[i]; 31 int ny = c + dy[i]; 32 if(judge(nx,ny)) 33 { 34 35 vis[nx][ny] = 1; 36 DFS(nx,ny,step+1); 37 vis[nx][ny] = 0; 38 } 39 } 40 } 41 42 int main() 43 { 44 int i, j, n, cas = 0; 45 scanf("%d",&n); 46 while(n--) 47 { 48 flag = 0; 49 scanf("%d%d",&p,&q); 50 memset(vis,0,sizeof(vis)); 51 vis[1][1] = 1; 52 DFS(1,1,1); 53 printf("Scenario #%d: ",++cas); 54 if(flag) 55 { 56 for(i = 1; i <= p * q; i++) 57 printf("%c%d",path[i][1] - 1 + 'A',path[i][0]); 58 } 59 else 60 printf("impossible"); 61 printf(" "); 62 if(n != 0) 63 printf(" "); 64 } 65 return 0; 66 }
来源：http://blog.csdn.net/lyhvoyage/article/details/18355471