poj-2488

A Knight's Journey

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 49498		Accepted: 16787

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

///深度搜索和回溯思想，还有字典顺序输出
///一个骑士，在p*q棋盘上以一个方向走两格，再垂直走一格的方式走遍每一个格子
#include<iostream>
#include<cstring>
using namespace std;
int vis[30][30],m,n,book[60][2];//vis用于标记走过的位置，book用于记录当前位置
int step[8][2]={{-1,-2},{1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}};//枚举出骑士走的方向,但要注意字典序，所以方向的陈列是有顺序的
bool flag=0;//作为找到答案的标志
void dfs(int x,int y,int t)
{

	if(t==(m*n))//判断是否所有格子均被遍历过
	{
		flag=1;
		return;
	}
	if(flag==1) return;  //因为是按照字典序进行查找的，一旦找到即为答案就直接返回
	for(int i=0;i<8;i++)   //遍历八个方向
		{
		int tx=x+step[i][0];//下一步x
		int ty=y+step[i][1];//下一步y（顺序很重要不要弄错）
		if(vis[tx][ty]==1)
		{
			vis[tx][ty]=0;     //走过的格子就不能再走，所以被标记为0
			book[t][0]=tx;  //记录当前格子
			book[t][1]=ty;
			dfs(tx,ty,t+1);    //继续搜索
			if(flag==1)  //若已找到答案，跳出循环
				break;
            else
                vis[tx][ty]=1;     //若此路不通，则将当前格子回溯（回溯过程的还原）
		}
	}
	return;
}
void print()
{
	for(int i=0;book[i][1]!=0;i++)
	{
		cout<<(char)(book[i][1]-1+'A')<<book[i][0];
	}
	cout<<endl;
	return;
}
int main()
{//i不能设为全局变量
	int x;
	cin>>x;
	for(int i=1;i<=x;i++)
	{
		flag=0;                //flvisg用于记录是否找到走完全图的最小字典序路径
		memset(vis,0,sizeof(vis));
		memset(book,0,sizeof(book));
		cin>>m>>n;
		for(int i=1;i<=m;i++)
			for(int j=1;j<=n;j++)
				vis[i][j]=1; //标记图中走的地方
		vis[1][1]=0;
		dfs(1,1,1);//3个参数，x，y为当前位置坐标，t为已经走过的格子数
		book[0][0]=1;
		book[0][1]=1;
		cout<<"Scenario #"<<i<<":"<<endl;
		if(flag==1)
		{
		  print();//输出结果
		}
		else
			cout<<"impossible"<<endl;//不存在就输出impossible
		cout<<endl;
	}

}