A Knight's Journey
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 29226   Accepted: 10023

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

深索水题。,,。主要是方向被规定了。

AC代码例如以下:

#include<iostream>
#include<cstring>
using namespace std; struct H
{
int x;
char y;
}b[30],c[30]; int dx[8]={-1,1,-2,2,-2,2,-1,1};
int dy[8]={-2,-2,-1,-1,1,1,2,2}; int a[30][30],vis[30][30];
int n,m,bj; void dfs(int h,int z,int cur)
{
int i;
if(cur==n*m)
{
if(bj==0)
{
for(i=0;i<cur;i++)
cout<<b[i].y<<b[i].x;
cout<<endl<<endl;
bj=1;
} }
else
{
for(i=0;i<8;i++)
{
int xx,yy;
xx=h+dx[i];
yy=z+dy[i];
if(!vis[xx][yy]&&xx>=1&&xx<=n&&yy>=1&&yy<=m)
{
vis[xx][yy]=1;
b[cur].x=xx;b[cur].y=(char)(yy+'A'-1);
dfs(xx,yy,cur+1);
vis[xx][yy]=0;
}
} }
} int main()
{
int t,cas=0;
cin>>t;
while(t--)
{
cas++;
memset(a,0,sizeof a);
memset(vis,0,sizeof vis);
cin>>n>>m;
bj=0;
b[0].x=1;b[0].y=1+'A'-1;
vis[1][1]=1;
cout<<"Scenario #"<<cas<<":"<<endl;
dfs(1,1,1);
if(bj==0)
{
cout<<"impossible"<<endl<<endl;
}
}
return 0;
}

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