POJ_2488: A Knight's Journey

POJ_2488: A Knight's Journey

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

题目：在一个n*m的格子里面，Knight从某一点出发，每次按照象棋中走“日”字的方法，判断他能否走完全部的格子。如果可以，按照字典序输出路径，如果不可以，输出”impossible“

分析：首先根据题目的意思，如果可以，Knight必定走完了所有的点，然后可以想到，我们从任意点出发，都能走完所有点，根据限制条件，按照字典序出发，于是我们可以从（1，1）点出发，有八个方向，考虑所有方向都可行并且按照字典序的话，方向的顺序应该为（-2，-1），（-2，1），（-1，-2），（-1，2），（1，-2）（1，2）（2，-1），（2，1），运用DFS，不断的递归求解，直至已走的点的数目等于n*m,说明有解，否则无解。在记录路径的时候，我考虑用两个数组来记录，一个记录X轴，一个记录Y轴，下标的变化过程跟已走的点数对应，并且需要用一个数组标记已走的点。具体贴代码吧~

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
using namespace std;
int dir[8][2]={-2,-1,-2,1,-1,-2,-1,2,1,-2,1,2,2,-1,2,1};
int id[28][28];
int n,m;
int flag;
int letter[28];
int number[28];
void dfs(int x,int y,int count1)
{
if(count1==n*m)
{
flag=true;
return ;
}
for(int k=0;k<8;k++)
{
int x1=x+dir[k][0];
int y1=y+dir[k][1];
if(x1>=1&&x1<=m&&y1>=1&&y1<=n&&id[x1][y1]==0)
{
count1++;
id[x1][y1]=1;
letter[count1]=x1;
number[count1]=y1;
if(count1==n*m)
{
flag=true;
return ;
}
dfs(x1,y1,count1);
if(flag!=true)
{
id[x1][y1]=0;
count1--;
}
}
}
}
int main()
{
int N;
scanf("%d",&N);
for(int h=1;h<=N;h++)
{
memset(id,0,sizeof(id));
scanf("%d%d",&n,&m);
flag=0;
memset(letter,0,sizeof(letter));
memset(number,0,sizeof(number));
id[1][1]=1;
letter[1]=1;
number[1]=1;
flag=false;
dfs(1,1,1);
int k=1;
printf("Scenario #%d:\n",h);
if(flag==true)
{
for(int i=1;i<=m*n;i++)
{
printf("%c%d",letter[i]-1+'A',number[i]);
}
printf("\n");
}
else
{
printf("impossible\n");
}
if(h!=N)
printf("\n");
}
return 0;
}