A Knight’s Journey
- 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
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 | #include<cstdio> #include<cstring> using namespace std; int array[30][30]; bool visited[30][30]; int p,q,n; bool checkfull(){ //?? for(int i=1;i<=p;i++){ for(int j=1;j<=q;j++){ if(visited[i][j]==false)return false; } } return true; } bool DFS(int i,int j,int cur){ if(i>0&&j>0&&i<=q&&j<=p&&(!visited[i][j])){ visited[i][j]=true; array[i][j]=cur; if(cur==p*q)return true; if(DFS(i-2,j-1,cur+1))return true; else if(DFS(i-2,j+1,cur+1))return true; else if(DFS(i-1,j-2,cur+1))return true; else if(DFS(i-1,j+2,cur+1))return true; else if(DFS(i+1,j-2,cur+1))return true; else if(DFS(i+1,j+2,cur+1))return true; else if(DFS(i+2,j-1,cur+1))return true; else if(DFS(i+2,j+1,cur+1))return true; visited[i][j]=false; array[i][j]=0; return false; }else return false; } bool dfs(){ for(int i=1;i<=q;i++){ for(int j=1;j<=p;j++){ if(DFS(i,j,1)){ return true; } } } return false; } void output(int k){ struct R{ int i; int j; }Result[30]; for(int i=1;i<=q;i++){ for(int j=1;j<=p;j++){ Result[array[i][j]].i=i; Result[array[i][j]].j=j; } } printf("Scenario #%d:\n",k); for(int i=1;i<=p*q;i++){ printf("%c%d",'A'+Result[i].i-1,Result[i].j); } printf("\n\n"); } int main(){ scanf("%d",&n); for(int i=1;i<=n;i++){ scanf("%d %d",&p,&q); memset(array,0,sizeof(array)); memset(visited,0,sizeof(visited)); if(dfs())output(i); else printf("Scenario #%d:\nimpossible\n\n",i); } } |