POJ 2488 A Knight's Journey
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 29226 | Accepted: 10023 |
Description

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
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
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
深索水题。,,。主要是方向被规定了。
AC代码例如以下:
#include<iostream>
#include<cstring>
using namespace std; struct H
{
int x;
char y;
}b[30],c[30]; int dx[8]={-1,1,-2,2,-2,2,-1,1};
int dy[8]={-2,-2,-1,-1,1,1,2,2}; int a[30][30],vis[30][30];
int n,m,bj; void dfs(int h,int z,int cur)
{
int i;
if(cur==n*m)
{
if(bj==0)
{
for(i=0;i<cur;i++)
cout<<b[i].y<<b[i].x;
cout<<endl<<endl;
bj=1;
} }
else
{
for(i=0;i<8;i++)
{
int xx,yy;
xx=h+dx[i];
yy=z+dy[i];
if(!vis[xx][yy]&&xx>=1&&xx<=n&&yy>=1&&yy<=m)
{
vis[xx][yy]=1;
b[cur].x=xx;b[cur].y=(char)(yy+'A'-1);
dfs(xx,yy,cur+1);
vis[xx][yy]=0;
}
} }
} int main()
{
int t,cas=0;
cin>>t;
while(t--)
{
cas++;
memset(a,0,sizeof a);
memset(vis,0,sizeof vis);
cin>>n>>m;
bj=0;
b[0].x=1;b[0].y=1+'A'-1;
vis[1][1]=1;
cout<<"Scenario #"<<cas<<":"<<endl;
dfs(1,1,1);
if(bj==0)
{
cout<<"impossible"<<endl<<endl;
}
}
return 0;
}
POJ 2488 A Knight's Journey的更多相关文章
- poj 2488 A Knight's Journey(dfs+字典序路径输出)
转载请注明出处:http://blog.csdn.net/u012860063?viewmode=contents 题目链接:http://poj.org/problem? id=2488 ----- ...
- pku 2488 A Knight's Journey (搜索 DFS)
A Knight's Journey Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 28697 Accepted: 98 ...
- 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 ...
- 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 ...
- POJ 2488-A Knight's Journey(DFS)
A Knight's Journey Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 31702 Accepted: 10 ...
- POJ 2488 A Knight's Journey(DFS)
A Knight's Journey Time Limit: 1000MSMemory Limit: 65536K Total Submissions: 34633Accepted: 11815 De ...
- 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) ...
- POJ 2488 A Knight's Journey (回溯法 | DFS)
题目链接:http://poj.org/problem?id=2488 题意: 在国际象棋的题盘上有一个骑士,骑士只能走“日”,即站在某一个位置,它可以往周围八个满足条件的格子上跳跃,现在给你一个p ...
- 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
- OptParse选项工具模块
OptParse是一个从Python2.3版本起引入的一个编写命令行工具模块,示例如下 ######example.py###### import optparse if __name__ == &q ...
- Android调用C#的WebService
Android调用C#写的WebService 学习自: http://www.cnblogs.com/kissazi2/p/3406662.html 运行环境 Win10 VS 2015 Andro ...
- leetcode 无重复字符的最长子串 python实现
这道题需要借助哈希查找key的O(n) 时间复杂度, 否则就会超时 初始化一个 哈希表\字典 dic 头指针start 初始为0 当前指针 cur 初始为0 最大长度变量 l 初始为0 用cur变量 ...
- [POI2015]Wilcze doły
[POI2015]Wilcze doły 题目大意: 给定一个长度为\(n(n\le2\times10^6)\)的数列\(A(1\le A_i\le10^9)\),可以从中选取不超过\(d\)个连续数 ...
- ASP.NET MVC HttpVerbs.Delete/Put Routes not firing
原文地址: https://weblog.west-wind.com/posts/2015/Apr/09/ASPNET-MVC-HttpVerbsDeletePut-Routes-not-firing ...
- hdu 1394 Minimum Inversion Number 逆序数/树状数组
Minimum Inversion Number Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showprob ...
- cookie、sessionStorage、localStorage 详解
转自--http://www.cnblogs.com/fly_dragon/p/3946012.html cookie Cookie的大小.格式.存储数据格式等限制,网站应用如果想在浏览器端存储用户的 ...
- Clever Little Box 电缆组件 USB A 插座 至 USB B 插头
http://china.rs-online.com/web/p/usb-cable-assemblies/7244156/ 产品详细信息 USB3.0适配器 superspeed USB将提供10x ...
- IP配制
http://blog.csdn.net/laoyiin/article/details/5722977
- 如何使用C#关键字const,readonly,static
如果有一个值不太会变化,我们经常使用const和readonly,这2者有何不同呢?有时候,我们也会在readonly之前加上关键字static,这又意味着什么呢? const ● const默认是静 ...