A Knight's Journey

Time Limit: 1000MS
Memory Limit: 65536K

Total Submissions: 34633
Accepted: 11815

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

题目简单翻译:

给你一个象棋中的马,一个n*m的棋盘,求是否能从一点出发,走遍整个棋盘,不重复走。如果能,按字典序输出第一个序列。如果不能,则输出“impossible”。

解题思路:

dfs,从一点出发,然而因为要字典序较小的,我们就选择(1,1)为起始点吧。注意延伸的方向,优先向字典序小的方向延伸。

代码:

#include<cstdio>
#include<cstring>
#include<queue> using namespace std;
int n,m;
int vis[][];
int dx[]={-,-,-,-,,,,};
int dy[]={-,,-,,-,,-,};
int a1[],a2[];
bool check(int x,int y)
{
return x>=&&x<n&&y>=&&y<m;
}
bool dfs(int x,int y,int depth)
{
if(depth==m*n)
{
for(int i=;i<depth;i++)
printf("%c%d",a1[i]+'A',a2[i]+);
puts("");
return true;
}
for(int i=;i<;i++)
{
int curx=x+dx[i];
int cury=y+dy[i];
if(check(curx,cury)&&vis[curx][cury]==)
{
a1[depth]=curx;
a2[depth]=cury;
vis[curx][cury]=;
if(dfs(curx,cury,depth+)) return true;
vis[curx][cury]=;
}
}
return false;
}
int main()
{
int T;
scanf("%d",&T);
int flag=;
while(T--)
{
if(flag) puts("");
scanf("%d%d",&m,&n);
memset(vis,,sizeof vis);
vis[][]=;
a1[]=,a2[]=;
printf("Scenario #%d:\n",++flag);
if(!dfs(,,)) puts("impossible");
}
return ;
}

POJ 2488 A Knight's Journey(DFS)的更多相关文章

  1. POJ 2488 A Knight's Journey (DFS)

    poj-2488 题意:一个人要走遍一个不大于8*8的国际棋盘,他只能走日字,要输出一条字典序最小的路径 题解: (1)题目上说的"The knight can start and end ...

  2. poj 2488 A Knight's Journey( dfs )

    题目:http://poj.org/problem?id=2488 题意: 给出一个国际棋盘的大小,判断马能否不重复的走过所有格,并记录下其中按字典序排列的第一种路径. #include <io ...

  3. POJ 2488 A Knight's Journey(深搜+回溯)

    A Knight's Journey Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other) ...

  4. POJ 2488 A Knight's Journey【DFS】

    补个很久之前的题解.... 题目链接: http://poj.org/problem?id=2488 题意: 马走"日"字,让你为他设计一条道路,走遍所有格,并输出字典序最小的一条 ...

  5. A Knight's Journey (DFS)

    题目: Background The knight is getting bored of seeing the same black and white squares again and agai ...

  6. POJ 2488 -- A Knight's Journey(骑士游历)

    POJ 2488 -- A Knight's Journey(骑士游历) 题意: 给出一个国际棋盘的大小,判断马能否不重复的走过所有格,并记录下其中按字典序排列的第一种路径. 经典的“骑士游历”问题 ...

  7. POJ2488-A Knight's Journey(DFS+回溯)

    题目链接:http://poj.org/problem?id=2488 A Knight's Journey Time Limit: 1000MS   Memory Limit: 65536K Tot ...

  8. POJ 2488-A Knight&#39;s Journey(DFS)

    A Knight's Journey Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 31702   Accepted: 10 ...

  9. poj 2488 A Knight's Journey 【骑士周游 dfs + 记忆路径】

    题目地址:http://poj.org/problem?id=2488 Sample Input 3 1 1 2 3 4 3 Sample Output Scenario #1: A1 Scenari ...

随机推荐

  1. angular--bootstrap实例日期控件【datepicker】

    head部分: <!--Bootstrap--> <link rel="stylesheet" href="/supProdom/script/boot ...

  2. RAC之RMAN恢复

    之前整理的RMAN 有关还原的文章: RMAN 系列(五) ---- RMAN 还原 与 恢复 http://blog.csdn.net/tianlesoftware/archive/2010/07/ ...

  3. Linux系统监控

    http://my.oschina.net/aiguozhe/blog/35730 http://my.oschina.net/aiguozhe/blog/35730

  4. Linux备份与恢复

    确定要备份的内容 在备份和还原系统时,Linux 基于文件的性质成了一个极大的优点.在 Windows 系统中,注册表与系统是非常相关的.配置和软件安装不仅仅是将文件放到系统上.因此,还原系统就需要有 ...

  5. Qt编程之实现在QFileDialog上添加自定义的widget

    上网搜索找到的方法如下: http://www.qtforum.org/article/20841/how-to-add-a-qwidget-in-qfiledialog.html#post78422 ...

  6. android UI之Shape详解_GradientDrawable

    在Android开发过程中,经常需要改变控件的默认样式, 那么通常会使用多个图片来解决.不过这种方式可能需要多个图片,比如一个按钮,需要点击时的式样图片,默认的式样图片. 这样就容易使apk变大. 那 ...

  7. 程序员求职之道(《程序员面试笔试宝典》)之求职有用网站及QQ群一览表

    技术学习网站 www.csdn.com www.iteye.com www.51cto.com http://www.cnblogs.com/ http://oj.leetcode.com/ http ...

  8. 自己写jstl标签解析long时间

    数据库里存储的是long型的时间,现在想输出到jsp页面,由于使用的是jstl标签,而要显示的是可读的时间类型,找来找去有个fmt:formatDate可以转化,但是只能转date型,long型则不可 ...

  9. poj 2096 Collecting Bugs(期望 dp 概率 推导 分类讨论)

    Description Ivan is fond of collecting. Unlike other people who collect post stamps, coins or other ...

  10. TypeScript 素描 - 模块解析、声明合并

    模块解析 模块解析有两种方式 相对方式  也就是以/或 ./或-/开头的,比如import jq  from "/jq" 非相对方式  比如 import model  from ...