A Knight's Journey

Time Limit: 1000MS	Memory Limit: 65536K
Total Submissions: 27258	Accepted: 9295

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

大致題意：給出一個p行q列的國際棋盤，馬可以從任意一個格子開始走，問馬能否不重複的走完所有的棋盤。如果可以，輸出按字典序排列最小的路徑。列印路徑時，列用大寫字母表示（A表示第一列），行用阿拉伯數字表示（從1開始），先輸出列，再輸出行。

分析：如果馬可以不重複的走完所有的棋盤，那麼它一定可以走到A1這個格子。所以我們只需從A1這個格子開始搜尋，就能保證字典序是小的；除了這個條件，我們還要控制好馬每次移動的方向，控制方向時保證字典序最小（即按照下圖中格子的序號搜尋）。控制好這兩個條件，直接從A1開始深搜就行了。

AC程式碼：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

int path[88][88], vis[88][88], p, q, cnt;
bool flag;

int dx[8] = {-1, 1, -2, 2, -2, 2, -1, 1};
int dy[8] = {-2, -2, -1, -1, 1, 1, 2, 2};

bool judge(int x, int y)
{
    if(x >= 1 && x <= p && y >= 1 && y <= q && !vis[x][y] && !flag)
        return true;
    return false;
}

void DFS(int r, int c, int step)
{
    path[step][0] = r;
    path[step][1] = c;
    if(step == p * q)
    {
        flag = true;
        return ;
    }
    for(int i = 0; i < 8; i++)
    {
        int nx = r + dx[i];
        int ny = c + dy[i];
        if(judge(nx,ny))
        {

            vis[nx][ny] = 1;
            DFS(nx,ny,step+1);
            vis[nx][ny] = 0;
        }
    }
}

int main()
{
    int i, j, n, cas = 0;
    scanf("%d",&n);
    while(n--)
    {
        flag = 0;
        scanf("%d%d",&p,&q);
        memset(vis,0,sizeof(vis));
        vis[1][1] = 1;
        DFS(1,1,1);
        printf("Scenario #%d:\n",++cas);
        if(flag)
        {
            for(i = 1; i <= p * q; i++)
                printf("%c%d",path[i][1] - 1 + 'A',path[i][0]);
        }
        else
            printf("impossible");
        printf("\n");
        if(n != 0)
            printf("\n");
    }
    return 0;
}