A Knight's Journey
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 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 Source
TUD Programming Contest 2005, Darmstadt, Germany
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#include <stdio.h> #include<string.h> int dx[8]= {-1,1,-2,2,-2,2,-1,1};//按字典顺序走必须这么写 int dy[8]= {-2,-2,-1,-1,1,1,2,2}; int visit[50][50],way[100][2];//visit记录是否访问。way记录路径 int p,q; int dfs(int x,int y,int r)//x,y记录当前所在位置。r记录下一步为第几步 { int nx,ny,i; if(r-1==p*q)//如果已经走完则输出 return 1; for(i=0; i<8; i++) { nx=x+dx[i]; ny=y+dy[i]; if(nx<1||nx>p||ny<1||ny>q||visit[nx][ny]) continue; visit[nx][ny]=1; way[r][0]=nx; way[r][1]=ny; if(dfs(nx,ny,r+1)) return 1; visit[nx][ny]=0; } return 0; } int main() { int t,i,k; scanf("%d",&t); for(k=1; k<=t; k++) { scanf("%d%d",&p,&q); memset(visit,0,sizeof visit); way[1][0]=way[1][1]=1; visit[1][1]=1; printf("Scenario #%d:\n",k); if(dfs(1,1,2)) { for(i=1; i<=p*q; i++) printf("%c%d",way[i][1]-1+'A',way[i][0]); printf("\n"); } else printf("impossible\n"); if(k!=t) printf("\n"); } return 0; }