computer(hdu2196)
Computer
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6378 Accepted Submission(s): 3211
Hint: the example input is corresponding to this graph. And from the graph, you can see that the computer 4 is farthest one from 1, so S1 = 3. Computer 4 and 5 are the farthest ones from 2, so S2 = 2. Computer 5 is the farthest one from 3, so S3 = 3. we also get S4 = 4, S5 = 4.
1 #include<iostream>
2 #include<string.h>
3 #include<algorithm>
4 #include<queue>
5 #include<math.h>
6 #include<stdlib.h>
7 #include<stack>
8 #include<stdio.h>
9 #include<ctype.h>
10 #include<map>
11 #include<vector>
12 #include<map>
13 using namespace std;
14 typedef struct acc
15 {
16 int id;
17 int val;
18 } ak;
19 vector<ak>vec[100005];
20 bool vis[100005];
21 typedef struct node
22 {
23 int id;
24 int mcost;
25 int scost;
26 int mid;
27 int sid;
28 node()
29 {
30 mcost = 0;
31 scost = 0;
32 mid = -1;
33 sid = -1;
34 }
35 } ss;
36 ss dp[100005];
37 int father[1000005];
38 void dfs1(int n);
39 void dfs2(int n);
40 int main(void)
41 {
42 int n,m;
43 while(scanf("%d",&n)!=EOF)
44 {
45 memset(father,-1,sizeof(father));
46 int i,j;
47 memset(dp,0,sizeof(dp));
48 for(i = 0; i < 100005; i++)
49 vec[i].clear();
50 for(i = 2; i <= n; i++)
51 {
52 int id,val;
53 scanf("%d %d",&id,&val);
54 ak a;
55 a.val = val;
56 a.id = i;
57 father[i] = id;
58 vec[id].push_back(a);
59 }
60 dfs1(1);
61 dfs2(1);
62 for(i = 1; i <= n; i++)
63 {
64 printf("%d\n",dp[i].mcost);
65 }
66 }
67 return 0;
68 }
69 void dfs1(int n)
70 {
71 for(int i = 0; i < vec[n].size(); i++)
72 {
73 ak d = vec[n][i];
74 {
75 dfs1(d.id);
76 if(dp[d.id].mcost+d.val > dp[n].mcost)
77 {
78 dp[n].scost = dp[n].mcost;
79 dp[n].sid = dp[n].mid;
80 dp[n].mcost = dp[d.id].mcost+d.val;
81 dp[n].mid = d.id;
82 }
83 else if(dp[d.id].mcost+d.val > dp[n].scost)
84 {
85 dp[n].scost = dp[d.id].mcost+d.val;
86 dp[n].sid = d.id;
87 }
88 }
89 }
90 }
91 void dfs2(int n)
92 {
93 for(int i = 0; i < vec[n].size(); i++)
94 {
95 ak d = vec[n][i];
96 if(dp[n].mid!=d.id)
97 {
98 if(dp[n].mcost + d.val > dp[d.id].mcost)
99 {
100 dp[d.id].scost = dp[d.id].mcost;
101 dp[d.id].sid = dp[d.id].mid;
102 dp[d.id].mcost = dp[n].mcost + d.val;
103 dp[d.id].mid = n;
104 }
105 else if(dp[n].mcost + d.val > dp[d.id].scost)
106 {
107 dp[d.id].scost = dp[n].mcost + d.val;
108 dp[d.id].sid = n;
109 }
110 }
111 else
112 {
113 if(dp[n].scost + d.val > dp[d.id].mcost)
114 {
115 dp[d.id].scost = dp[d.id].mcost;
116 dp[d.id].sid = dp[d.id].mid;
117 dp[d.id].mcost = dp[n].scost + d.val;
118 dp[d.id].mid = n;
119 }
120 else if(dp[n].scost + d.val > dp[d.id].scost)
121 {
122 dp[d.id].scost = dp[n].scost + d.val;
123 dp[d.id].sid = n;
124 }
125 }
126 dfs2(d.id);
127 }
128 }
computer(hdu2196)的更多相关文章
- Computer(HDU2196+树形dp+树的直径)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2196 题目: 题意:有n台电脑,每台电脑连接其他电脑,第i行(包括第一行的n)连接u,长度为w,问你每 ...
- [HDU2196]Computer(DP)
传送门 题意 给出一棵树,求离每个节点最远的点的距离 思路 对于我这种菜鸡,真是难啊. 每个点的距离它最远的点,除了在它子树中的,还有在它子树之外的,所以这几个状态都得表示出来. 我们能够很简单的求出 ...
- MySQL数据库的安装与配置(windows)
MySQL是目前最为流行的开放源码的数据库,是完全网络化的跨平台的关系型数据库系统,它是由瑞典MySQLAB公司开发,目前属于Oracle公司.任何人都能从Internet下载MySQL软件,而无需支 ...
- 【FSFA 读书笔记】Ch 2 Computer Foundatinons(2)
Hard Disk Technology 1. 机械硬盘内部构造 几个重要概念:Sector(扇区),Head(读写头),Track(磁道),Cylinder(柱面). 如果一个文件比较大,磁盘的写入 ...
- hdu 2196(Computer 树形dp)
A school bought the first computer some time ago(so this computer's id is 1). During the recent year ...
- Multiple View Geometry in Computer Vision Second Edition by Richard Hartley 读书笔记(二)
// Chapter 2介绍的是2d下的投影变换,摘录下了以下定理 Result 2.1. The point x lies on the line l if and only if xTl = 0. ...
- 【FSFA 读书笔记】Ch 2 Computer Foundatinons(1)
Data Organization 1. 进制转换. 按照正常的书写顺序写一个数字(无论多少进制),其中最左边的列称为“最高有效符号”,最右边的列称为“最低有效符号”. (The right-most ...
- 图像分类(三)GoogLenet Inception_v3:Rethinking the Inception Architecture for Computer Vision
Inception V3网络(注意,不是module了,而是network,包含多种Inception modules)主要是在V2基础上进行的改进,特点如下: 将滤波器尺寸(Filter Size) ...
- 【Network architecture】Rethinking the Inception Architecture for Computer Vision(inception-v3)论文解析
目录 0. paper link 1. Overview 2. Four General Design Principles 3. Factorizing Convolutions with Larg ...
随机推荐
- Python异步IO之select
1. select模块的基本使用(以socket为例) 1 # -*- coding:utf-8 -*- 2 # Author:Wong Du 3 4 import select 5 import s ...
- (转载)SQL Server 2008 连接JDBC详细图文教程
点评:SQL Server 2008是目前windows上使用最多的sql数据库,2008的安装机制是基于framework重写的,特点是非常耗时间SQL Server 2008是目前windows上 ...
- 《手把手教你》系列技巧篇(四十七)-java+ selenium自动化测试-判断元素是否显示(详解教程)
1.简介 webdriver有三种判断元素状态的方法,分别是isEnabled,isSelected 和 isDisplayed,其中isSelected在前面的内容中已经简单的介绍了,isSelec ...
- academy
academy at/in school都行,academy一般用at. The word comes from the Academy in ancient Greece, which derive ...
- acquire, acre, across
acquire An acquired taste is an appreciation [鉴赏] for something unlikely to be enjoyed by a person w ...
- flink02------1.自定义source 2. StreamingSink 3 Time 4窗口 5 watermark
1.自定义sink 在flink中,sink负责最终数据的输出.使用DataStream实例中的addSink方法,传入自定义的sink类 定义一个printSink(),使得其打印显示的是真正的ta ...
- 零基础学习java------20---------反射
1. 反射和动态代理 参考博文:https://blog.csdn.net/sinat_38259539/article/details/71799078 1.0 什么是Class: 我们都知道,对象 ...
- TLSv1.3 Support:主流 Web 客户端和服务端对 TLSv1.3 的支持情况
TLSv1.3 Support:主流 Web 客户端和服务端对 TLSv1.3 的支持情况 请访问原文链接:https://sysin.org/blog/tlsv1-3-support/,查看最新版. ...
- [php代码审计] Typecho 1.1 -反序列化Cookie数据进行前台Getshell
环境搭建 源码下载:https://github.com/typecho/typecho/archive/v1.1-15.5.12-beta.zip 下载后部署到web根目录,然后进行安装即可,其中注 ...
- springmvc中的异常处理方法
//1.自定义异常处理类 2.编写异常处理器 3.配置异常处理器 package com.hope.exception;/** * 异常处理类 * @author newcityma ...