poj 2488 A Knight's Journey

 A Knight's Journey

 

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
impossible
Scenario #2:
A1B3C1A2B4C2A3B1C3A4B2C4
Scenario #3:

题目大意：有一个p行q列的棋盘这个棋子可以向这八个方向走，问这个棋子是否可以走完棋盘所有的地方如果能走完输出路径，否则输出impossible，输出的路径要按字典序排列。

因为输出路径时要按顺序，所以我们按下图标记的顺序走就可以。

#include<cstdio>#include<cstring>#include<iostream>using namespace std;int p,q,vis[15][15],flag,dis[30][2];int dir[8][2]={-1,-2,//可以走的八个方向1,-2,-2,-1,2,-1,-2,1,2,1,-1,2,1,2};bool judge(int x, int y){      if(x>=1&&x<=p&&y>=1&&y<=q&&!vis[x][y]&&!flag)          return true;      return false;  } void dfs(int x,int y,int step){dis[step][0]=x;dis[step][1]=y;if(step==p*q){flag=1;return ;}for(int i=0;i<8;i++){int xx=x+dir[i][0];int yy=y+dir[i][1];if(judge(xx,yy)){vis[xx][yy]=1;dfs(xx,yy,step+1);vis[xx][yy]=0;}}}int main(){int T,i;scanf("%d",&T);for(i=1;i<=T;i++){memset(vis,0,sizeof(vis));memset(dis,0,sizeof(dis));vis[1][1]=1;flag=0;scanf("%d%d",&p,&q);printf("Scenario #%d:\n",i);dfs(1,1,1);if(flag){for(int j=1;j<=p*q;j++)printf("%c%d",dis[j][1]-1+'A',dis[j][0]);putchar('\n');}elseprintf("impossible\n");if(i!=T)putchar('\n');}return 0;}