A Knight's Journey

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这道题目关键的就是字典序(题目要求字母要尽量小其次数字要尽量小,所以就是按照左上、右上、左下、右下的顺序)这就需要dir数组要有一个固定的顺序,还有就是输入时是先输入列再输入行

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
#include <stdio.h>  #include <string.h>   int go[8][2]= {{-1,-2},{1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}};int vis[30][30];  int n,m,flag;  struct node  {      int x,y;  } a[30];void dfs(int x,int y,int step)  {      a[step].x=x,a[step].y=y;    if(step==n*m)    {          for(int i=1; i<=step; i++)              printf("%c%d",a[i].y-1+'A',a[i].x);        printf("\n");          flag=1;      }      if(flag)return;      for(int i=0; i<8; i++)      {          int xx=x+go[i][0];          int yy=y+go[i][1];          if(xx>0&&xx<=n&&yy>0&&yy<=m&&vis[xx][yy]==0)         {              vis[xx][yy]=1;              dfs(xx,yy,step+1);              vis[xx][yy]=0;        }      }  }  int main()  {      int t,ci=1;      scanf("%d",&t);      while(t--)      {          memset(vis,0,sizeof(vis));          flag=0;          scanf("%d%d",&n,&m);          printf("Scenario #%d:\n",ci++);          vis[1][1]=1;         dfs(1,1,1);        if(flag==0)printf("impossible\n");          printf("\n");      }      return 0;  }