poj 2488

A Knight's Journey

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 30099 Accepted: 10320

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

TUD Programming Contest 2005, Darmstadt, Germany

AC代码：

#include<iostream>#include<cstring>using namespace std;int p,q;int G[30][30];int sucess;char ch[1000];int a[8][2]={{-1,-2},{1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}};int dfs(int x,int y,int step){    if(x<0 || x>p-1 || y<0 || y>q-1 || G[x][y]) return 0;    G[x][y]=1;    if(step==p*q){        ch[2*step]='\0';        return 1;    }    for(int k=0;k<8;k++){        if(dfs(x+a[k][0],y+a[k][1],step+1)){            ch[2*(step+1)-2]=y+a[k][1]+'A';            ch[2*(step+1)-1]=x+a[k][0]+'1';            return 1;        }    }    G[x][y]=0;    return 0;}int main(){    int T; cin>>T;    for(int cas=1;cas<=T;cas++){        cin>>p>>q;        for(int j=0;j<q;j++){            for(int i=0;i<p;i++){                sucess=0;                memset(G,0,sizeof(G));                ch[0]=j+'A';                ch[1]=i+'1';                if(dfs(i,j,1)){                    sucess=1;                    break;                }            }            if(sucess)                break;        }        cout<<"Scenario #"<<cas<<":"<<endl;        if(!sucess)            cout<<"impossible"<<endl<<endl;        else            cout<<ch<<endl<<endl;    }    return 0;}