搜索 B

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

31 12 34 3

Sample Output

Scenario #1:A1Scenario #2:impossibleScenario #3:A1B3C1A2B4C2A3B1C3A4B2C4

题意：这个马的遍历方式是 f[8][2]={{-2,-1},{-2,1},{-1,-2},{-1,2},{1,-2},{1,2},{2,-1},{2,1}} 八个方向，如果全能遍历，那么找出按字典序排列最小的序列。

如果要保证按字典序排列最小那就从头开始 深搜 。

#if 0#include<iostream>#include<cstring>using namespace std;#define max 30int ax[max],ay[max],q,p;bool vis[max][max],flag;int f[8][2]={{-2,-1},{-2,1},{-1,-2},{-1,2},{1,-2},{1,2},{2,-1},{2,1}};//方向int judge(int x,int y)                                             //判断有没有越界{ if(x>=0 && x<q && y>=0 && y<p && vis[q][p]==0)  return 1;  else  return 0; }void dfs(int a,int b,int t)             //a是x b是y t是遍历的个数{ ax[t]=a; ay[t]=b; vis[a][b]=1;  if(t==p*q-1)                 //如果全都访问一遍了 那就返回 {    flag=1;                      //标记       return ;}  for(int i=0; i<8; i++) {int dx=ax[t]+f[i][0];      //int dy=ay[t]+f[i][1];if(judge(dx,dy) && !vis[dx][dy]){  vis[dx][dy]=1;      dfs(dx,dy,t+1);  if(flag)                     return ;  vis[dx][dy]=0; //回溯}} return ;}int main(){int n,ans=1;cin>>n;while(n--){ flag=0; memset(vis,false,sizeof(vis));          //每一次都要初始化 memset(ax,0,sizeof(ax)); memset(ay,0,sizeof(ay));    cin>>p>>q; cout<<"Scenario #"<<ans<<":"<<endl;    ans++;   for(int i=0; i<q; i++)  {    for(int j=0; j<p; j++)  {  dfs(i,j,0);  if(flag)   break;  }  if(flag)  break;  }    if(flag)  {    for(int i=0; i<q*p; i++)   cout<<char(ax[i]+'A')<<ay[i]+1;   cout<<endl<<endl;  }  else   cout<<"impossible"<<endl<<endl;  } } #endif