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. 学界| UC Berkeley提出新型分布式框架Ray:实时动态学习的开端—— AI 应用的系统需求:支持(a)异质、并行计算,(b)动态任务图,(c)高吞吐量和低延迟的调度,以及(d)透明的容错性。

    学界| UC Berkeley提出新型分布式框架Ray:实时动态学习的开端 from:https://baijia.baidu.com/s?id=1587367874517247282&wfr ...

  2. B1270 [BeijingWc2008]雷涛的小猫 dp

    这个题的原始方法谁都会,但是n^3会T.之后直接优化,特别简单,就是每次处理出来每层的最大值,而不用枚举.之前没这么做是因为觉得在同一棵树的时候没有下落,所以不能用这个方法.后来想明白了,在同一棵树上 ...

  3. 二维矩阵相乘 in C++

    #include <iostream> #include <vector> #include <string> #include <sstream> # ...

  4. SQL Server2012 T-SQL基础教程--读书笔记(5-7章)

    SQL Server2012 T-SQL基础教程--读书笔记(5-7章) SqlServer T-SQL 示例数据库:点我 Chapter 05 表表达式 5.1 派生表 5.1.1 分配列别名 5. ...

  5. 3.4 存储简单数据的利器——Preferences

    <?xml version="1.0" encoding="utf-8"?> <manifest xmlns:android="ht ...

  6. Python 33(2)进程理论

    一:什么是进程         进程指的是一个正在进行 / 运行的程序,进程是用来描述程序执行过程的虚拟概念 进程vs程序 程序:一堆代码 进程:程序的执行的过程 进程的概念起源于操作系统,进程是操作 ...

  7. Oracle创建用户教程

    计算机-->管理-->应用程序与服务-->(OracleOraDb11g_home1TNSListener 和 OracleServiceORCL 服务)->启动服务 打开Or ...

  8. Java同步容器总结

    <0>StringBuffer适用于多线程场景,StringBuilder适用于字符串拼接[堆栈封闭] `Vector`实现`List`接口,底层和`ArrayList`类似,但是`Vec ...

  9. POJ 1160 DP

    题目: poj 1160 题意: 给你n个村庄和它的坐标,现在要在其中一些村庄建m个邮局,想要村庄到最近的邮局距离之和最近. 分析: 这道题.很经典的dp dp[i][j]表示建第i个邮局,覆盖到第j ...

  10. HBase的集群搭建(1、3、5节点都适用)

    见 5 hbase-shell + hbase的java api