POJ2488——A Knight's Journey

A Knight's Journey Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 41519 Accepted: 14125

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

Source

TUD Programming Contest 2005, Darmstadt, Germany

题目大意：

有T组测试数据，每组有 p 和 q 两个数，表示棋盘大小，p 行 q 列，棋子能否不重复的走完棋盘中所有的格子，是则按照字典序输出路径，否则输出 impossible；

思路：

就是一道规规矩矩的深搜 的题目，按照字典序遍历，所以周围八个可走格子的遍历顺序为：

#include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> using namespace std; int dir[8][2] = {{-1,-2},{1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}}; //方向数组 int vis[30][30]; //标记是否走过 int mp[30][2]; //记录路径 int p, q; int flag; //标记是否走完 bool out(int x, int y) //判断是否越界 { if( x>=0&&x<p&&y>=0&&y<q ) return true; return false; } void dfs(int x, int y, int deep) { mp[deep][0] = x; mp[deep][1] = y; if( deep==p*q ) //遍历完了flag改变，跳出 { flag = 1; return; } for( int i = 0;i < 8;i++ ) { int xx = x + dir[i][0]; int yy = y + dir[i][1]; if( !out(xx,yy) ) continue; //若越界，继续下一步 if( !flag &&!vis[xx][yy] ) { vis[xx][yy] = 1; dfs(xx,yy,deep+1); vis[xx][yy] = 0; } } } int main() { int T; cin>>T; for( int test = 1;test <= T; test++ ) { cin>>p>>q; memset(vis,0,sizeof(vis)); flag = 0; vis[0][0] = 1; dfs(0,0,1); printf("Scenario #%d:\n",test); if( flag ) { for( int i = 1;i <= p*q;i++ ) { printf("%c%d",mp[i][1]+'A',mp[i][0]+1); } cout<<endl; } else cout<<"impossible"<<endl; cout<<endl; } return 0; }