A Knight’s Journey
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 34085 Accepted: 11621
Description
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
知识点：DFS
题意：找到第一条走遍棋盘的路径，并且输出路径。
难点：扩展状态，按‘日’字形扩展。
初始状态：从A1开始，个数为1；
扩展方式：按走‘日’字形8个方向；
目标状态：棋盘全部遍历到
1 #include<cstdlib> 2 #include<cstdio> 3 #include<iostream> 4 #include<algorithm> 5 #include<cstring> 6 using namespace std; 7 char map1[30][30]; 8 int map2[30][30]; 9 int vis[30][30]; 10 int t,p,q,flag; 11 int dx[] = {-1,1,-2,2,-2,2,-1,1}; 12 int dy[] = {-2,-2,-1,-1,1,1,2,2}; 13 void dfs(int x,int y,int cnt) 14 { 15 if(cnt==p*q&&!flag) 16 { 17 flag=1; 18 for(int i=0;i<cnt;i++) 19 printf("%c%d",map1[i][0],map2[i][0]); 20 printf(" "); 21 return; 22 } 23 if(x<0||x>=p||y<0||y>=q) 24 return; 25 for(int i=0;i<8;i++) 26 { 27 int nx=x+dx[i]; 28 int ny=y+dy[i]; 29 //printf("%d @@@%d %d ",ny,nx,cnt); 30 if(!vis[nx][ny]&&nx>=0&&nx<p&&ny>=0&&ny<q) 31 { 32 map1[cnt][0]='A'+ny; 33 map2[cnt][0]=nx+1; 34 vis[nx][ny]=1; 35 // printf("%d %d %d ",ny,nx,cnt); 36 dfs(nx,ny,cnt+1); 37 vis[nx][ny]=0; 38 } 39 } 40 41 } 42 int main() 43 { 44 scanf("%d",&t); 45 int ha=0; 46 while(t--) 47 { 48 scanf("%d%d",&p,&q); 49 memset(vis,0,sizeof(vis)); 50 flag=0; 51 printf("Scenario #%d: ",++ha); 52 for(int i=0;i<p;i++) 53 { 54 for(int j=0;j<q;j++) 55 { 56 if(flag==0) 57 { 58 map1[0][0]='A'+i; 59 map2[0][0]=0+j+1; 60 vis[i][j]=1; 61 dfs(i,j,1); 62 vis[i][j]=0; 63 } 64 else 65 break; 66 } 67 if(flag==1) 68 break; 69 } 70 71 if(flag==0) 72 printf("impossible "); 73 } 74 75 return 0; 76 } 77 //3 78 //1 1 79 //2 3 80 //4 3
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