A Knight's Journey（深度搜索）

A Knight's Journey

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 42686 Accepted: 14506

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

Source

#include <iostream>#include <cstdio>#include <cstring>using namespace std;int vis[10][10];int s[10][10]={{-2,-1},{-2,1},{-1,-2},{-1,2},{1,-2},{1,2},{2,-1},{2,1}};//按字典序输出，x从小到大int x1[30],y1[30];int p,q,flag,n;void dfs(int x,int y,int step){   x1[step]=x;   y1[step]=y;   if(step==p*q)   {      flag=1;      return ;   }   for(int i=0;i<8;i++)   {       int tx=x+s[i][0];       int ty=y+s[i][1];       if(1<=tx&&tx<=q&&1<=ty&&ty<=p&&!flag&&!vis[tx][ty])//注意p和q       {          vis[tx][ty]=1;          dfs(tx,ty,step+1);          vis[tx][ty]=0;       }   }  return ;}int main(){    scanf("%d",&n);        for(int k=1;k<=n;k++)    {        flag=0;        scanf("%d%d",&p,&q);        memset(vis,0,sizeof(vis));        vis[1][1]=1;        dfs(1,1,1);        printf("Scenario #%d:\n",k);        if(flag)        {           for(int i=1;i<=p*q;i++)           {               printf("%c%d",x1[i]-1+'A',y1[i]);           }        }        else        printf("impossible");        if(k==n)        printf("\n");        else        printf("\n\n");//注意格式    }    return 0;}