H

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
题目意思:
在一个给定大小的国际象棋的棋盘,上面有一个马,问马可以走完棋盘吗?

解法:

#include <iostream>
#include <string.h>
using namespace std;

const int MAX = 300 +10;
int dx[8]={-2,-2,-1,-1,1,1,2,2};
int dy[8]={-1,1,-2,2,-2,2,-1,1};
int L,W;
int Map[MAX][MAX];
int flag;
int str1[MAX],str2[MAX];

void print()
{
    for(int i = 1;i <= L;i++)
        for(int j = 1;j <= W;j++)
        {
            str1[Map[i][j] ] = i;
            str2[Map[i][j] ] = j;
        }
}

void Dfs (int x,int y,int count1)
{
    if(count1 == L*W)
    {
        flag = 1;
        print();
        return;
    }
    for(int i = 0;i < 8;i++)
    {
        if(flag) return;
        else
        {
            int x1 = x+dx[i];
            int y1 = y+dy[i];
            if(x1>=1 && x1<=L && y1>=1 && y1<=W && !Map[x1][y1] )
            {
       //         cout<<"x1 = "<<x1<<" y1 = "<<y1<<" count = "<<count1<<endl;
                Map[x1][y1] = count1 + 1;
                Dfs(x1,y1,count1+1);
                Map[x1][y1] = 0;
            }
        }
    }

    return ;
}

int main()
{
    int n;
    cin>>n;
    int n0 = 0;
    while(n--)
    {
        cin>>W>>L;
        cout<<"Scenario #"<<++n0<<":"<<endl;
        memset(Map,0,sizeof(Map));
        flag = 0;
        Map[1][1] = 1;
        Dfs(1,1,1);

        if(flag == 0)
            cout<<"impossible"<<endl;
        else
        {
            for(int i = 1;i <= L*W;i++)
            {
                char temp = str1[i] +'A'-1;
                cout<<temp<<str2[i];
            }
            cout<<endl;
        }
        cout<<endl;
    }

    return 0;
}