A Knight's Journey_DFS

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

【题意】给出m*n的方阵（但输入时先输入的是n，再输入m），问马是否能走遍棋盘，输出字典序的第一种路径。

【思路】用mp[i][0]表示第i步所在那个格子的横坐标，用mp[i][1]表示第i步所在那个格子的纵坐标。

字典序的话，注意di数组的顺序。用一个dfs就好啦。

#include <iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
const int N=100;
int vis[N][N];
int mp[N][2];
int n,m;
bool flag;
int di[8][2]={-2,-1,-2,1,-1,-2,-1,2,1,-2,1,2,2,-1,2,1};
bool go(int x,int y)
{
    if(x<0||x>=n||y<0||y>=m) return false;
    else return true;
}

void dfs(int i,int j,int k)
{
    if(k==n*m)
    {
        for(int i=0;i<k;i++)
        {
            printf("%c%d",mp[i][0]+'A',mp[i][1]+1);
        }
        printf("
");
        flag=true;
       // return ;
    }
    else
    for(int x=0;x<8;x++)
    {
        int xx=i+di[x][0];
        int yy=j+di[x][1];
        if(!vis[xx][yy]&&go(xx,yy)&&!flag)
        {
            vis[xx][yy]=1;
            mp[k][0]=xx;
            mp[k][1]=yy;
            dfs(xx,yy,k+1);
            vis[xx][yy]=0;

        }
    }
}



int main()
{
  int t,cas=1;
  scanf("%d",&t);
  while(t--)
  {
      scanf("%d%d",&m,&n);
      memset(vis,0,sizeof(vis));
      vis[0][0]=1;
      mp[0][0]=0;
      mp[0][1]=0;
      flag=false;
      printf("Scenario #%d:
",cas++);
      dfs(0,0,1);

      if(!flag) printf("impossible
");
    puts("");
  }
    return 0;
}