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
- myeclipse svn重新定位 本地文件 svn 重新定位
我们在用工具myeclipse开发项目时,当资源库存储空间不够时,我们就需要添加资源库,涉及到我们切换项目资源库,下面就介绍一下svn资源库重新定位步骤 1,window到show view到othe ...
- paper 167:GPU的使用Theano之tutorial
Theano之使用GPU 英文版本:http://deeplearning.net/software/theano/tutorial/using_gpu.html using the ...
- codeforces 111A/112C Petya and Inequiations
题目:Petya and Inequiations传送门: http://codeforces.com/problemset/problem/111/A http://codeforces.com/p ...
- 关于设置shadowPath的重要性
这是超级容易添加阴影到iOS中的任何视图.所有您需要做的是 添加QuartzCore框架到项目中(如果不存在的话) 导入QuartzCore到您的执行文件 添加一行如[myView.layer set ...
- (转)运行jar应用程序引用其他jar包的四种方法 -- ClassLoader应用
转:http://longdick.iteye.com/blog/332580 大家都知道一个java应用项目可以打包成一个jar,当然你必须指定一个拥有main函数的main class作为你这个j ...
- vijos 1054 牛场围栏 【想法题】
这题刚看完后第一个想到的方法是背包 但仔细分析数据范围后会发现这题用背包做复杂度很高 比如对于这样的数据 2 100 2999 2898 (如果有神犇可以用背包过掉这样的数据 请回复下背包的做法) - ...
- HP Loadrunner 12.53的下载与安装
HP Loadrunner 12.53的下载与安装 HP Loadrunner 12.53的下载:http://pan.baidu.com/s/1c23axHm HP Loadrunner 12. ...
- 用vim写go代码——vim-go插件
GoImport:导入包 GoImport!:导入远程包 GoImportAs: 导入包并且重命名
- poj3468 A Simple Problem with Integers (树状数组做法)
题目传送门 A Simple Problem with Integers Time Limit: 5000MS Memory Limit: 131072K Total Submissions: 1 ...
- java虚拟机规范(se8)——class文件格式(四)
4.7 属性 属性用于class文件格式中的ClassFile,field_info,method_info和Code_attribute结构. 所有的属性都是下面的格式: attribute_inf ...