POJ 2488 A Knight's Journey

D - A Knight's Journey

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

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

题意：给出一个规格为8 * 8的棋盘，一个骑士可以从任意一个格子开始走，并且只能走“日”，问骑士能否不重复的走完整个棋盘。如果能，输出按字典序排列最小的路径，若不能，输出“impossible"。

分析：假设骑士可以不重复的走完整个棋盘，那么他一定可以走到A1这个格子，所以我们只需从A1开始搜索，并且控制好骑士移动的方向，得到的一定是字典序最小的路径。控制好条件后，就直接从A1进行深搜就可以了，记得将搜过的地方进行标记。
<pre name="code" class="cpp">#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>

using namespace std;

int flag,m,n;
int visit[30][30];
int v[30][2];
//搜索方向
int go[8][2] = {{-2,-1},{-2,1},{-1,-2},{-1,2},{1,-2},{1,2},{2,-1},{2,1}};

//打印
void print(int sum)
{
    for(int i = 0;i < sum;i++)
    {
        printf("%c%d",v[i][0] - 1 + 'A',v[i][1]);
    }
    printf("\n");
}

void dfs(int x,int y,int sum)
{
    if(flag)    //已经搜到答案，结束搜索
        return ;
    if(sum == n * m)  //搜到答案
    {
        flag = 1;
        print(sum);
        return ;
    }
    else
    {
        //对8个方向进行搜素
        for(int i = 0;i < 8;i++)
        {
            //移动之后的位置记为(nx,ny)
            int nx = x + go[i][0];
            int ny = y + go[i][1];
            //判断是否可以移动以及是否访问过
            if(1 <= nx && nx <= n && 1 <= ny && ny <= m && visit[nx][ny] == 0)
            {
                visit[nx][ny] = 1;//标记为1
                v[sum][0] = nx;  //记录位置
                v[sum][1] = ny;
                dfs(nx,ny,sum + 1); //搜索下一个位置
                visit[nx][ny] = 0;//取消标记
            }
        }
    }
}

int main()
{
   int t;
   int cont = 0;
   scanf("%d",&t);
   while(t--)
   {
       memset(visit,0,sizeof(visit));//初始化
       memset(v,0,sizeof(v));
       scanf("%d %d",&m,&n);
       flag = 0;
       visit[1][1] = 1;
       v[0][0] = 1;
       v[0][1] = 1;
       if(cont++ > 0)
            printf("\n");
       printf("Scenario #%d:\n",cont);
       dfs(1,1,1);
       if(flag == 0)
            printf("impossible\n");
   }
   return 0;
}