POJ 2488 A Knight's Journey

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8546		Accepted: 2894

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

1.按照字典序搜索
即有：
int dir[8][2]={
	{-2,-1},{-2,1},
	{-1,-2},{-1,2},
	{1,-2},{1,2},
	{2,-1},{2,1}
	};
这个字典顺序牢记。。。

2.可直接从(1,1)搜索，不成功，则impossible(对于本题可以这样,理论上没做分析呵呵~)

3.看到其他discussion,说要放大数组，不过我一开始就够大了,没考虑

#include<stdio.h>
#include<string.h>
bool visit[50][50],flag;
int b[8]={-2,-2,-1,-1,1,1,2,2},a[8]={-1,1,-2,2,-2,2,-1,1};
int dx[65],dy[65],n,x,y,count1;
void dfs(int sum,int xx,int yy)
{
 int i;
 if(sum==x*y)
 {
  for(i=1;i<=sum;i++)
  {
   printf("%c",dy[i]-1+'A');
   printf("%d",dx[i]);
  }
  printf("/n");
  flag=true;
  return;
 }
 for(i=0;i<8&&!flag;i++)
 {
  if(xx+a[i]>0&&xx+a[i]<=x&&yy+b[i]>0&yy+b[i]<=y&&!visit[xx+a[i]][yy+b[i]]&&!flag)
  {
   dx[sum+1]=xx+a[i];
   dy[sum+1]=yy+b[i];
   visit[xx+a[i]][yy+b[i]]=true;
   dfs(sum+1,xx+a[i],yy+b[i]);
   visit[xx+a[i]][yy+b[i]]=false;
  }
 }
} 
int main()
{
 freopen("in.txt","r",stdin);
 freopen("out.txt","w",stdout);
 while(scanf("%d",&n)!=EOF)
 {
  count1=0;
  while(n--)
  {
   flag=false;
   printf("Scenario #%d:/n",++count1);
   memset(visit,false,sizeof(visit));
   scanf("%d%d",&x,&y);
   dy[1]=1;
   dx[1]=1;
   visit[1][1]=true;
   dfs(1,1,1);
   if(!flag)  printf("impossible/n");
   printf("/n");
  }
 }
}