A Knight's Journey

A Knight's Journey

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 43095 Accepted: 14619

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

#include<stdio.h>#include<iostream>#include<algorithm>#include<string.h>#include<stdlib.h>#include<time.h>#include<math.h>#include<queue>;#include<stack>;#include <iomanip>;using namespace std;int t,b[100][2],v[10][10],n,m,p;int dir[8][2]= {{-1,-2},{1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}};int go(int x,int y){    if(1<=x&&x<=n&&1<=y&&y<=m&&!v[x][y]&&!p)        return 1;    return 0;}void dfs(int x,int y,int c){    b[c][0]=y;    b[c][1]=x;    if(c==n*m)    {        p=1;        return;    }    for(int i=0; i<8; i++)    {        int ex=x+dir[i][0];        int ey=y+dir[i][1];        if(go(ex,ey))        {            v[ex][ey]=1;            dfs(ex,ey,c+1);            v[ex][ey]=0;        }    }}int main(){    scanf("%d",&t);    for(int i=1; i<=t; i++)    {         p=0;        scanf("%d%d",&n,&m);        memset(v,0,sizeof(v));        printf("Scenario #%d:\n",i);        v[1][1]=1;        dfs(1,1,1);        if(p==1)        {            for(int i=1; i<=n*m; i++)            {                printf("%c%d",b[i][0]+'A'-1,b[i][1]);            }        }        else            printf("impossible");        printf("\n");        if(i!=t)            printf("\n");    }}