A Knight's Journey(DFS)深搜

时间：2015-04-14 23:22:14 收藏：0 阅读：262

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

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;
}

原文：http://blog.csdn.net/u012965373/article/details/45047745

A Knight&#39;s Journey(DFS)深搜

A Knight's Journey(DFS)深搜