Poj2488 -- A Knight & #39; s Journey (dfs, server guard problem), poj2488 -- adfs

Poj2488 -- A Knight's Journey (dfs, server guard problem), poj2488 -- adfs
A Knight's Journey

Time Limit:1000 MS		Memory Limit:65536 K
Total Submissions:31147		Accepted:10655

Description

Background 
The knight is getting bored of seeing the same black and white squares 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 between different square numbers 1 ,..., p exist, q describes how many different square letters exist. these are the first q letters of the Latin alphabet: ,...

Output

The output for every scenario begins with a line ining "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 shoshould 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 shoshould 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 provides a matrix of n columns in the m row, and asks if the server guard can finish all vertices. Each vertex can only go once and requires the minimum output in the Lexicographic Order. If you can finish the process, you can start from A1. The Lexicographic Order is the minimum as long as dfsA1 starts. Check whether the process can be completed. Pay attention to the search order.

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int t , n , m , sum ;int mm[30][30] ;int pre[1000] ;int a[8][2] = { {-2,-1},{-2,1},{-1,-2},{-1,2},{1,-2},{1,2},{2,-1},{2,1} };int dfs(int x,int y,int temp){    if( temp == sum )        return 1 ;    int flag = 0 , xx , yy , i ;    for(i = 0 ; i < 8 ; i++)    {        xx = x + a[i][0] ;        yy = y + a[i][1] ;        if( xx >= 0 && xx < n && yy >= 0 && yy < m && !mm[xx][yy] )        {            mm[xx][yy] = 1 ;            pre[x*m+y] = xx*m+yy ;            flag = dfs(xx,yy,temp+1);            if( flag ) return flag ;            mm[xx][yy] = 0 ;        }    }    return flag ;}int main(){    int i , j , k , tt ;    scanf("%d", &t);    for(tt = 1 ; tt <= t ; tt++)    {        memset(mm,0,sizeof(mm));        memset(pre,-1,sizeof(pre));        scanf("%d %d", &m, &n);        sum = n*m ;        mm[0][0] = 1 ;        k = dfs(0,0,1);        printf("Scenario #%d:\n", tt);        if(k == 0)            printf("impossible");        else        {            for(i = 0 , k = 0 ; i < sum ; i++)            {                printf("%c%c", k/m+'A', k%m+'1');                k = pre[k] ;            }        }        printf("\n\n");    }    return 0;}