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Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

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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

经典的DFS，大概题意就是让骑士走遍棋盘的所有格子。一路走到黑，看这条路能不能使骑士把格子都走完。

#include<cstdio>
#include<string.h>
using namespace std;
int vis[30][30];
char str[2000];
int s[8][2]={-2,-1,-2,1,-1,-2,-1,2,1,-2,1,2,2,-1,2,1};//这种处理方式很常用的，要记住，即骑士当前所在位置的周围所能走到的位置
int p,q;
int dfs(int x,int y,int sum,int cnt)
{
    if(sum==p*q) return 1;//sum是记录走过的格子
    int x1,y1;
    for(int i=0;i<8;i++)
    {
        x1=x+s[i][0];
        y1=y+s[i][1];
        if(x1>=0&&x1<q&&y1>=0&&y1<p&&!vis[x1][y1])//边界条件及判断是否访问过吗
        {
            vis[x1][y1]=1;
            str[cnt+1]=x1+'A';//开一个str数组，用来记录
            str[cnt+2]=y1+'1';
            if(dfs(x1,y1,sum+1,cnt+2))//记得这里要写成sum+1,cnt+2
                return 1;
            vis[x1][y1]=0;//回溯
        }
    }
    return 0;
}
int main()
{
    int t;
    scanf("%d",&t);
    for(int i=1;i<=t;i++)
    {
        scanf("%d %d",&p,&q);
        memset(vis,0,sizeof(vis));
        memset(str,0,sizeof(str));
        str[0]='A';
        str[1]='1';
        vis[0][0]=1;
        if(dfs(0,0,1,1)){
            printf("Scenario #%d:\n",i);
            for(int j=0;j<strlen(str);j++)
                printf("%c",str[j]);
            printf("\n\n");//记得每组数据输出后，有个空行
        }
        else{
            printf("Scenario #%d:\n",i);
            printf("impossible\n\n");
        }
    }
    return 0;
}