UVALive 3902 Network (树+dfs)
Consider a tree network with n nodes where the internal nodes correspond to servers and the terminal
nodes correspond to clients. The nodes are numbered from 1 to n. Among the servers, there is an
original server S which provides VOD (Video On Demand) service. To ensure the quality of service for
the clients, the distance from each client to the VOD server S should not exceed a certain value k. The
distance from a node u to a node v in the tree is dened to be the number of edges on the path from u
to v. If there is a nonempty subset C of clients such that the distance from each u in C to S is greater
than k , then replicas of the VOD system have to be placed in some servers so that the distance from
each client to the nearest VOD server (the original VOD system or its replica) is k or less.
Given a tree network, a server S which has VOD system, and a positive integer k, nd the minimum
number of replicas necessary so that each client is within distance k from the nearest server which has
the original VOD system or its replica.
For example, consider the following tree network.
In the above tree, the set of clients is f1, 6, 7, 8, 9, 10, 11, 13g, the set of servers is f2, 3, 4, 5, 12,
14g, and the original VOD server is located at node 12.
For k = 2, the quality of service is not guaranteed with one VOD server at node 12 because the
clients in f6, 7, 8, 9, 10g are away from VOD server at distance > k. Therefore, we need one or more
replicas. When one replica is placed at node 4, the distance from each client to the nearest server of
f12, 4g is less than or equal to 2. The minimum number of the needed replicas is one for this example.
Input
Your program is to read the input from standard input. The input consists of T test cases. The number
of test cases (T) is given in the rst line of the input. The rst line of each test case contains an integer
n (3 n 1; 000) which is the number of nodes of the tree network. The next line contains two
integers s (1 s n) and k (k 1) where s is the VOD server and k is the distance value for ensuring
the quality of service. In the following n
UVALive 3902 Network (树+dfs)的更多相关文章
- [UVALive 3902] Network
图片加载可能有点慢,请跳过题面先看题解,谢谢 一道简单的贪心题,而且根节点已经给你了(\(S\)),这就很好做了. 显然,深度小于等于 \(k\) 的都不用管了(\(S\) 深度为0),那么我们只需要 ...
- UVaLive 3902 Network (无根树转有根树,贪心)
题意:一个树形网络,叶子是客户端,其他的是服务器.现在只有一台服务器提供服务,使得不超k的客户端流畅,但是其他的就不行了, 现在要在其他结点上安装服务器,使得所有的客户端都能流畅,问最少要几台. 析: ...
- UVALive3902 Network[贪心 DFS&&BFS]
UVALive - 3902 Network Consider a tree network with n nodes where the internal nodes correspond to s ...
- HDU 5692 线段树+dfs序
Snacks Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Sub ...
- Tsinsen A1505. 树(张闻涛) 倍增LCA,可持久化线段树,DFS序
题目:http://www.tsinsen.com/A1505 A1505. 树(张闻涛) 时间限制:1.0s 内存限制:512.0MB 总提交次数:196 AC次数:65 平均分: ...
- 51 nod 1681 公共祖先 (主席树+dfs序)
1681 公共祖先 基准时间限制:1 秒 空间限制:131072 KB 分值: 80 难度:5级算法题 有一个庞大的家族,共n人.已知这n个人的祖辈关系正好形成树形结构(即父亲向儿子连边). 在另 ...
- BZOJ_3252_攻略_线段树+dfs序
BZOJ_3252_攻略_线段树+dfs序 Description 题目简述:树版[k取方格数] 众所周知,桂木桂马是攻略之神,开启攻略之神模式后,他可以同时攻略k部游戏.今天他得到了一款新游戏< ...
- 【XSY2534】【BZOJ4817】树点涂色 LCT 倍增 线段树 dfs序
题目大意 Bob有一棵\(n\)个点的有根树,其中\(1\)号点是根节点.Bob在每个点上涂了颜色,并且每个点上的颜色不同.定义一条路径的权值是:这条路径上的点(包括起点和终点)共有多少种不同的颜 ...
- 【BZOJ】3991: [SDOI2015]寻宝游戏 虚树+DFS序+set
[题意]给定n个点的带边权树,对于树上存在的若干特殊点,要求任选一个点开始将所有特殊点走遍后返回.现在初始没有特殊点,m次操作每次增加或减少一个特殊点,求每次操作后的总代价.n,m<=10^5. ...
随机推荐
- 聊聊Dubbo - Dubbo可扩展机制实战
1. Dubbo的扩展机制 在Dubbo的官网上,Dubbo描述自己是一个高性能的RPC框架.今天我想聊聊Dubbo的另一个很棒的特性, 就是它的可扩展性. 如同罗马不是一天建成的,任何系统都一定是从 ...
- 安装ThinkPHP
ThinkPHP5的环境要求如下: PHP >= 5.4.0 PDO PHP Extension MBstring PHP Extension CURL PHP Extension 严格来说,T ...
- Java常用数据结构Set, Map, List
1. Set Set相对于List.Map是最简单的一种集合.集合中的对象不按特定的方式排序,并且没有重复对象. 特点: 它不允许出现重复元素: 不保证和政集合中元素的顺序 允许包含值为null的元素 ...
- EAM(Enterprise Asset Management)企业资产管理系统
EAM (Enterprise Asset Management)的缩写,EAM系统是指企业资产管理系统. EAM系统是在资产比重较大的企业,在资产建设.维护中减少维护成本,提高资产运营效率,通过现代 ...
- RHEL/CentOS通用性能优化、安全配置参考
RHEL/CentOS通用性能优化.安全配置参考 本文的配置参数是笔者在实际生产环境中反复实践总结的结果,完全适用绝大多数通用的高负载.安全性要求的网络服务器环境.故可以放心使用. 若有异议,欢迎联系 ...
- 前端每日实战:38# 视频演示如何用纯 CSS 创作阶梯文字特效
效果预览 按下右侧的"点击预览"按钮可以在当前页面预览,点击链接可以全屏预览. https://codepen.io/comehope/pen/MXYBEM 可交互视频教程 此视频 ...
- 使用xorm将结构体转为sql文件
操作步骤 (1)定义结构体 type User struct { Id int //表id Name string //姓名 ...}12345(2)编写代码,执行自动增量同步(mysql为例) im ...
- 项目搭建(三):自定义DLL
说明:程序中有些自定义的控件类型在TestStack.White框架中没有涉及,需要引入自定义的DLL,通过鼠标点击事件处理 使用:将自定义的ClassLibrary2.dll拷贝到项目/bin/de ...
- PHP解码unicode编码的中文字符
问题背景:晚上在抓取某网站数据,结果在数据包中发现了这么一串编码的数据:"......\u65b0\u6d6a\u5fae\u535a......www.jinyuanbao.cn" ...
- Java反射实现Servlet处理多个请求--server分发
import javax.servlet.ServletException; import javax.servlet.annotation.WebServlet; import javax.serv ...