A Knight's Journey

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 188   Accepted Submission(s) : 40

Problem 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. <br>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：给出一个棋盘的规格，问你能否在不重复的情况下吧每一个方格都走一遍，按照国际象棋的走法，东南西北只走日，按字典序输出
思路：广搜记录路径
注意横坐标是字母，纵坐标数字，按照字典序（ABC），以及搜索及时停止，输出，基本问题不大
Source Code
#include<iostream>#include<stdio.h>#include<string.h>using namespace std;struct chess{    int x;    int y;    };int dx[]={-1,1,-2,2,-2,2,-1,1};int dy[]={-2,-2,-1,-1,1,1,2,2};bool vis[101][101];chess ans[101];char zimubiao[]={'A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z'};int flag;int n,m; int judge(int x,int y) {     if(x>0&&x<=n&&y>0&&y<=m)     return 1;     return 0;     }void dfs(int x,int y,int num){    int i,j,nextx,nexty;    ans[num].x=x;    ans[num].y=y;    if(flag)return ;    if(num==n*m)    {        for(i=1;i<=n*m;i++)            {                cout<<zimubiao[ans[i].y-1]<<ans[i].x;                }            cout<<endl;        flag=1;        return ;    }    if(flag)return ;    for(i=0;i<8;i++)    {        nextx=x+dx[i];        nexty=y+dy[i];        if(judge(nextx,nexty)==1&&vis[nextx][nexty]==0)        {            vis[nextx][nexty]=1;            dfs(nextx,nexty,num+1);            vis[nextx][nexty]=0;            }        }    }int main(){    int t;    int c=1,i,j;    cin>>t;    for(;c<=t;c++)    {        cin>>n>>m;        memset(vis,0,sizeof(vis));        flag=0;        vis[1][1]=1;        cout<<"Scenario #"<<c<<":"<<endl;        dfs(1,1,1);        if(flag==0)cout<<"impossible"<<endl;        if(c<=t)cout<<endl;        }    return 0;    }