poj2488--A Knight's Journey(dfs，骑士问题)

A Knight's Journey

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

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

3
1 1
2 3
4 3

Sample Output

Scenario #1:
A1

Scenario #2:
impossible

Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4

Source

TUD Programming Contest 2005, Darmstadt, Germany

给出m行n列的棋盘，问骑士能不能走全然部的点，每一个点仅仅能走一次，要求字典序最小输出。

假设能走完。那么一定能够从A1開始走。字典序最小 仅仅要dfsA1開始，看能不能走完，注意搜索的顺序。

#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;
}