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
- CentOS7.4 关闭firewall防火墙,改用iptables
1.关闭默认的firewall防火墙 systemctl stop firewalld.service #停止firewall systemctl disable firewalld.service ...
- 【贪心】Google Code Jam Round 1A 2018 Waffle Choppers
题意:给你一个矩阵,有些点是黑的,让你横切h刀,纵切v刀,问你是否能让切出的所有子矩阵的黑点数量相等. 设黑点总数为sum,sum必须能整除(h+1),进而sum/(h+1)必须能整除(v+1). 先 ...
- 【原创】实战padding oracle漏洞
首先关于padding oracle漏洞的原理请看: 步入正传~~ 搭建漏洞利用环境Perl 环境下载地址:链接:http://pan.baidu.com/s/1skFxVm1 密码:anuw 首先查 ...
- hdu 5316 Magician(2015多校第三场第1题)线段树单点更新+区间合并
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5316 题意:给你n个点,m个操作,每次操作有3个整数t,a,b,t表示操作类型,当t=1时讲a点的值改 ...
- opencv第三课 Canny边缘检测
#include<stdio.h> #include<iostream> #include<opencv2\opencv.hpp> using namespace ...
- HDU 5137 How Many Maos Does the Guanxi Worth 最短路 dijkstra
How Many Maos Does the Guanxi Worth Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 512000/5 ...
- 《学习OpenCv》 笔记(1)
P1-P17 废话 可跳过 不过讲了如何搭建环境,如果你没有搭建的话,可以查看我的另外一个博文,详细讲了如何构建OpenCv的编程环境 P19 开始编写第一个代码
- flex socket policy
@ flex的as3代码是具备使用origin tcp socket通信能力的. @ 如果是flex builder本机调试,那么可以直连tcp的server. @ 如果flex发布在webserve ...
- rac 10g 10.2.0.1升级到10.2.0.5具体解释
RAC 10.2.0.1 升级到 10.2.0.5 一. 准备: Patch 包:p8202632_10205_LINUX.zip 节点数:3个节点 RAC1 RAC2 ...
- SpringMVC和Springboot的区别(网摘)
spring boot 我理解就是把 spring spring mvc spring data jpa 等等的一些常用的常用的基础框架组合起来,提供默认的配置,然后提供可插拔的设计,就是各种 sta ...