poj2488(DFS)之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 
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
题目大意：给你p*q的棋盘，问你马能否以其中一格为起点遍历整个棋盘？如果可以，按字典序输出路径
思路表：打表字典序，若存在路径，则从A1开始即可遍历整个棋盘
AC代码如下：
<pre name="code" class="cpp">#include <iostream>  #include <cstdio>  #include <cstring>    using namespace std;    int p,q,flag1;  int coun,cnt;  int a[30][30];  char s[1000];  int dy[8] = {-2, -2, -1, -1, 1, 1, 2, 2};  int dx[8] = {-1, 1, -2, 2, -2, 2, -1, 1};    void dfs(int row,int col,int coun,int cnt)  {      if(row<1 || row>p || col<1 || col>q) return;      if(cnt==p*q*2 && coun==p*q)      {          for(int i=0; i<cnt; i++)          {              printf("%c",s[i]);          }          cout<<endl<<endl;          flag1=1;          return;      }      a[row][col]=coun;      for(int i=0; i<8 && flag1==0; i++)      {          if(row+dx[i]>=1 && row+dx[i]<=p && col+dy[i]>=1 && col+dy[i]<=q && a[row+dx[i]][col+dy[i]]==0)          {              a[row+dx[i]][col+dy[i]]=1;                s[cnt]='A'+col+dy[i]-1;              s[cnt+1]='1'+row+dx[i]-1;                dfs(row+dx[i],col+dy[i],coun+1,cnt+2);                a[row+dx[i]][col+dy[i]]=0;          }      }  }    int main()  {      int Case;      cin>>Case;      for(int flag=1; flag<=Case; flag++)      {          flag1=cnt=coun=0;          memset(a,0,sizeof(a));            cin>>p>>q;          printf("Scenario #%d:\n",flag);            s[cnt]='A';          s[cnt+1]='1';          dfs(1,1,coun+1,cnt+2);          if(flag1==0) cout<<"impossible"<<endl<<endl;      }      return 0;  }