A Knight's Journey(DFS)深搜

A Knight's Journey

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 33634		Accepted: 11450

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 

Source

TUD Programming Contest 2005, Darmstadt, Germany

题意：

背景
骑士越来越无聊的一次又一次看到相同的黑白方块,并决定旅行
世界各地。当骑士移动,它是在一个方向和一个广场两个正方形垂直于这一点。骑士是棋盘的世界他是生活在。我们的骑士生活的棋盘上小面积比普通8 * 8,但它仍然是矩形。你能帮助这个冒险的骑士做旅行计划吗?

找到一个路径,骑士访问每各方格。骑士可以在任何方格开始和结束。

输入输出解释：

首先输入一个n代表的代表一个有几组测试用例。

接下来每组的输入为一个p, q

代表着小于8 * 8 的p * q的棋盘。

问你能不能遍历整个棋盘的每一个方格，如果能输出路径，不能则输出impossible. 

对于这道题目来说，刚开始看没有理解是怎么走的，因为正常的思维都是在线上走，那么这里我们可以把格子虚拟成为线，

那么画出图来，就是在线上走的了，这样会好理解的多。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#define MAXN 50
using namespace std;
int x_move[8] = {-2, -2, -1, -1, 1, 1, 2, 2};    //优先向字典序小的方向搜索，有点贪心，呵呵  
int y_move[8] = {-1, 1, -2, 2, -2, 2, -1, 1};  
bool Map[30][30], flag;  
int r, c;  
char res[100];  
void dfs (int x, int y, int step, int k)  
{  
    if (flag)  
        return ;  
    res[k] = x + 'A', res[k+1] = y + '1';    //记录路径  
    if (step == r * c)    //走完全部格子，说明成功  
    {  
        flag = true;  
        return ;  
    }  
    int i, tx, ty;  
    for (i = 0; i < 8; i++)  
    {  
        tx = x + x_move[i];  
        ty = y + y_move[i];  
        if (tx >= 0 && ty >= 0 && tx < r && ty < c)  
        {  
            if (!Map[tx][ty])  
            {  
                Map[tx][ty] = true;  
                dfs (tx, ty, step+1, k+2);  
                Map[tx][ty] = false;  
            }  
        }  
    }  
}  
  
int main()  
{  
    int t, k = 1;  
    cin >> t;  
    while (t--)  
    {  
        cin >> c >> r;  
        memset (Map, false, sizeof(Map));  
        Map[0][0] = true;  
        flag = false;  
        dfs (0, 0, 1, 0);  
        cout << "Scenario #" << k++ << ":\n";  
        if (!flag)  
            cout << "impossible\n";  
        else  
        {  
            res[2*r*c] = 0;    //封好字符串  
            cout << res << endl;  
        }  
        cout << endl;  
    }  
    return 0;  
}