POJ2488:A Knight's Journey

A Knight's Journey

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 39700		Accepted: 13487

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
 
思路：dfs生成全排列遍历，注意起点与终点是任意的。因为要求输出是字典序，注意跳跃时的顺序。

#include <cstdio>
#include <cstring>
using namespace std;
const int MAXN=30;
int n;
int p,q;
int dy[8]={-1,1,-2,2,-2,2,-1,1};
int dx[8]={-2,-2,-1,-1,1,1,2,2};
int vis[MAXN][MAXN];
bool mark;
void dfs(int dep,int y,int x,char s[])
{
    if(mark)
        return ;
    if(dep==2*p*q)
    {
        mark=true;
        s[dep]=0;
        printf("%s
",s);
        return ;
    }
    for(int i=0;i<8;i++)
    {
        int ny=y+dy[i];
        int nx=x+dx[i];
        if(0<=ny&&ny<p&&0<=nx&&nx<q&&!vis[ny][nx])
        {
            vis[ny][nx]=1;
            s[dep]='A'+nx;
            s[dep+1]='1'+ny;
            dfs(dep+2,ny,nx,s);
            vis[ny][nx]=0;
        }
    }
}
int main()
{
    scanf("%d",&n);
    for(int cas=1;cas<=n;cas++)
    {
        scanf("%d%d",&p,&q);
        printf("Scenario #%d:
",cas);
        
        mark=false;
        for(int i=0;i<p;i++)
        {
            for(int j=0;j<q;j++)
            {
                memset(vis,0,sizeof(vis));
                char s[60]="";
                s[0]='A'+i;
                s[1]='1'+j;
                vis[i][j]=1;
                dfs(2,i,j,s);
                if(mark)
                    break;
            }
            if(mark)    break;
        }
        if(!mark)
            printf("impossible
");
        printf("
");
    }
    return 0;
}