A Knight's Journey POJ - 2488

A Knight's Journey

 POJ - 2488 

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

题解+解析：

#include <iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include <cstring>
#include <string>
#define INF 0x7f7f7f7f
typedef long long ll;
using namespace std;
bool visit[100][100];
bool flag;//标记是否已完成遍历。
int p = 1;
int m, n;
int t;
int move[8][2] = {-2,-1,-2,1,-1,-2,-1,2,1,-2,1,2,2,-1,2,1};//移动位置的数组，顺序不能变，因为题目要求字典序。
char str[10000];
void dfs(int x, int y, int c)
{
	if (flag)//已完成，直接返回
	{
		return;
	}
	str[c] = x + 'A';//将该位置写入数组
	str[c+1] = y + '1';
	c += 2;//游标移动
	if (c == m*n*2)//完成遍历则打印路径
	{
		flag = 1;
		printf("Scenario #%d:\n", p);
		printf("%s\n", str);
		return;
	}
	for (int i = 0; i < 8; i++)//遍历八种跳法
	{
		int nx = x + move[i][0], ny = y + move[i][1];
		if (0<=nx&&nx<n&&0<=ny&&ny<m&&!visit[nx][ny])//符合条件，深搜该位置
		{
			visit[nx][ny] = true;
			dfs(nx, ny, c);
			visit[nx][ny] = false;//回溯
		}
	}
	c -= 2;//回溯
	str[c] = 0;
	str[c+1] = 0;
}
int main()
{
	
	scanf("%d", &t);
	while (t--)
	{
		for (int i = 0; i < 200; i++)
		{
			str[i] = 0;
		}
		scanf("%d%d",&m,&n);
		flag = 0;
		for (int i = 0; i < n; i++)//从任意地点开始，遍历整个图（其实不用，只要从A1开始搜就可以了，因为如果该棋盘有解，必定经过A1，所以不需要遍历图。此处写麻烦了。）
		{
			for (int j = 0; j < m; j++)
			{
				if (flag)
				{
					break;
				}
				else
				{
					memset(visit, false, sizeof(visit));//初始化访问数组。
					visit[i][j] = true;//将i，j位置设为true
					dfs(i, j, 0);//从i，j开始搜索
				}
			}
		}
		if (!flag)
		{
			printf("Scenario #%d:\n", p);
			printf("impossible\n");
		}
		if (t)//注意换行符的打印
		{
			printf("\n");
		}
		p++;
	}
	return 0;
}