Channel Allocation
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 15546   Accepted: 7871

Description

When a radio station is broadcasting over a very large area, repeaters are used to retransmit the signal so that every receiver has a strong signal. However, the channels used by each repeater must be carefully chosen so that nearby repeaters do not interfere with one another. This condition is satisfied if adjacent repeaters use different channels.

Since the radio frequency spectrum is a precious resource, the number of channels required by a given network of repeaters should be minimised. You have to write a program that reads in a description of a repeater network and determines the minimum number of channels required.

Input

The input consists of a number of maps of repeater networks. Each map begins with a line containing the number of repeaters. This is between 1 and 26, and the repeaters are referred to by consecutive upper-case letters of the alphabet starting with A. For example, ten repeaters would have the names A,B,C,...,I and J. A network with zero repeaters indicates the end of input.

Following the number of repeaters is a list of adjacency relationships. Each line has the form:

A:BCDH

which indicates that the repeaters B, C, D and H are adjacent to the repeater A. The first line describes those adjacent to repeater A, the second those adjacent to B, and so on for all of the repeaters. If a repeater is not adjacent to any other, its line has the form

A:

The repeaters are listed in alphabetical order.

Note that the adjacency is a symmetric relationship; if A is adjacent to B, then B is necessarily adjacent to A. Also, since the repeaters lie in a plane, the graph formed by connecting adjacent repeaters does not have any line segments that cross.

Output

For each map (except the final one with no repeaters), print a line containing the minumum number of channels needed so that no adjacent channels interfere. The sample output shows the format of this line. Take care that channels is in the singular form when only one channel is required.

Sample Input

2
A:
B:
4
A:BC
B:ACD
C:ABD
D:BC
4
A:BCD
B:ACD
C:ABD
D:ABC
0

Sample Output

1 channel needed.
3 channels needed.
4 channels needed.

POJ 1129 Channel Allocation DFS 回溯的更多相关文章

  1. poj 1129 Channel Allocation ( dfs )

    题目:http://poj.org/problem?id=1129 题意:求最小m,使平面图能染成m色,相邻两块不同色由四色定理可知顶点最多需要4种颜色即可.我们于是从1开始试到3即可. #inclu ...

  2. 迭代加深搜索 POJ 1129 Channel Allocation

    POJ 1129 Channel Allocation Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 14191   Acc ...

  3. POJ 1129 Channel Allocation(DFS)

    Channel Allocation Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 13173   Accepted: 67 ...

  4. POJ 1129 Channel Allocation 四色定理dfs

    题目: http://poj.org/problem?id=1129 开始没读懂题,看discuss的做法,都是循环枚举的,很麻烦.然后我就决定dfs,调试了半天终于0ms A了. #include ...

  5. poj 1129 Channel Allocation(图着色,DFS)

    题意: N个中继站,相邻的中继站频道不得相同,问最少需要几个频道. 输入输出: Sample Input 2 A: B: 4 A:BC B:ACD C:ABD D:BC 4 A:BCD B:ACD C ...

  6. poj 1129 Channel Allocation

    http://poj.org/problem?id=1129 import java.util.*; import java.math.*; public class Main { public st ...

  7. poj 1154 letters (dfs回溯)

    http://poj.org/problem?id=1154 #include<iostream> using namespace std; ]={},s,r,sum=,s1=; ][]; ...

  8. PKU 1129 Channel Allocation(染色问题||搜索+剪枝)

    题目大意建模: 一个有N个节点的无向图,要求对每个节点进行染色,使得相邻两个节点颜色都不同,问最少需要多少种颜色? 那么题目就变成了一个经典的图的染色问题 例如:N=7 A:BCDEFG B:ACDE ...

  9. POJ 1321 棋盘问题 (DFS + 回溯)

    题目链接:http://poj.org/problem?id=1321 题意:中文题目,就不多说了...... 思路: 解题方法挺多,刚开始想的是先从N行中选择出来含有“#”的K行,再在这K行中放置K ...

随机推荐

  1. C# winform listBox中的项上下移动(转)

    C# winform listBox中的项上下移动 分类: C# winform2009-06-24 12:37 876人阅读 评论(0) 收藏 举报 winformc#object //上移节点   ...

  2. hdu 1242(BFS+优先队列)

    Rescue Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Subm ...

  3. WPF:通过Window.DataContext实现窗口间传值

    通过Window.DataContext实现窗口之间的传值,特别是跨窗口控件的联动,具有无可比拟的优势.实现方法如下: 1.  MainWindow.xaml,在Window.DataContext中 ...

  4. leetcode树相关

    目录 144前序遍历 94中序遍历(98验证二叉搜索树.230二叉搜索树中第K小的元素) 145后序遍历 102/107层次遍历(104二叉树最大深度.103 105从前序与中序遍历序列构造二叉树 1 ...

  5. 0428-mysql(事务、权限)

    1.事务 “事务”是一种可以保证“多条语句一次性执行完成”或“一条都不执行”的机制. 两种开始事务的方法: 1.set  autocommit = 0; //false,此时不再是一条语句一个事务了, ...

  6. go函数初级

    一.简介 在go语言中,函数的功能是非常强大的,以至于被认为拥有函数式编程语言的多种特性. 二.介绍 1.一个程序中包含了很多的函数:函数式基本的代码块 2.函数编写的顺序是无关紧要的:鉴于可读性的需 ...

  7. ROS-节点-Topic

    前言:本部分主要介绍ros一些基础功能的使用,包括创建和编译工作空间.功能包.节点以及话题. 第一种方式:使用roboware studio软件操作 1.1 创建工作空间 回车然后点击保存. 1.2 ...

  8. 常用MIME类型(Flv,Mp4的mime类型设置)

    也许你会在纳闷,为什么我上传了flv或MP4文件到服务器,可输入正确地址通过http协议来访问总是出现“无法找到该页”的404错误呢?这就表明mp4格式文件是服务器无法识别的,其实,这是没有在iis中 ...

  9. Cannot find module 'crc'

    这个时候你只需要打开你nodejs安装的目录,在其中执行 npm install crc(这里查什么模块(module)就安装什么模块).

  10. 联想笋尖S90(S90-t 、S90-u)解锁BootLoader

    工具下载链接: http://pan.baidu.com/s/1eSgZuka 备用下载链接: http://pan.baidu.com/s/1dFKqSId 本篇教程,仅限于联想笋尖S90(S90- ...