POJ 2488 A Knight's Journey

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;
}