Helvetic Coding Contest 2017 online mirror (teams allowed, unrated) J
Description
Heidi's friend Jenny is asking Heidi to deliver an important letter to one of their common friends. Since Jenny is Irish, Heidi thinks that this might be a prank. More precisely, she suspects that the message she is asked to deliver states: "Send the fool further!", and upon reading it the recipient will ask Heidi to deliver the same message to yet another friend (that the recipient has in common with Heidi), and so on.
Heidi believes that her friends want to avoid awkward situations, so she will not be made to visit the same person (including Jenny) twice. She also knows how much it costs to travel between any two of her friends who know each other. She wants to know: what is the maximal amount of money she will waste on travel if it really is a prank?
Heidi's n friends are labeled 0 through n - 1, and their network of connections forms a tree. In other words, every two of her friends a, b know each other, possibly indirectly (there is a sequence of friends starting from a and ending on b and such that each two consecutive friends in the sequence know each other directly), and there are exactly n - 1 pairs of friends who know each other directly.
Jenny is given the number 0.
The first line of the input contains the number of friends n (3 ≤ n ≤ 100). The next n - 1 lines each contain three space-separated integers u, vand c (0 ≤ u, v ≤ n - 1, 1 ≤ c ≤ 104), meaning that u and v are friends (know each other directly) and the cost for travelling between u and vis c.
It is guaranteed that the social network of the input forms a tree.
Output a single integer – the maximum sum of costs.
4
0 1 4
0 2 2
2 3 3
5
6
1 2 3
0 2 100
1 4 2
0 3 7
3 5 10
105
11
1 0 1664
2 0 881
3 2 4670
4 2 1555
5 1 1870
6 2 1265
7 2 288
8 7 2266
9 2 1536
10 6 3378
5551
In the second example, the worst-case scenario goes like this: Jenny sends Heidi to the friend labeled by number 2 (incurring a cost of 100), then friend 2 sends her to friend 1 (costing Heidi 3), and finally friend 1 relays her to friend 4 (incurring an additional cost of 2).
题意:通俗一点,求树的每条分支的和,把最大的拿出来
解法:dfs,遍历到叶子就比较一下喽
#include <bits/stdc++.h>
using namespace std;
int n;
vector<pair<int,int>>q[];
int sum;
int vis[];
void dfs(int x,int cot,int pre)
{
if(vis[x])
{
return;
}
vis[x]=;
for(int i=; i<q[x].size(); i++)
{
if(vis[q[x][i].first])
{
{
sum=max(sum,cot);
continue;
}
}
dfs(q[x][i].first,cot+q[x][i].second,x);
}
}
int main()
{
cin>>n;
for(int i=; i<n; i++)
{
int v,u,c;
cin>>v>>u>>c;
q[v].push_back({u,c});
q[u].push_back({v,c});
}
dfs(,,);
cout<<sum<<endl;
return ;
}
Helvetic Coding Contest 2017 online mirror (teams allowed, unrated) J的更多相关文章
- Helvetic Coding Contest 2017 online mirror (teams allowed, unrated)
G. Fake News (easy) time limit per test 1 second memory limit per test 256 megabytes input standard ...
- Helvetic Coding Contest 2017 online mirror (teams allowed, unrated) M
Description The marmots have prepared a very easy problem for this year's HC2 – this one. It involve ...
- Helvetic Coding Contest 2017 online mirror (teams allowed, unrated) A
Description Your search for Heidi is over – you finally found her at a library, dressed up as a huma ...
- Helvetic Coding Contest 2019 online mirror (teams allowed, unrated)
http://codeforces.com/contest/1184 A1 找一对整数,使x^x+2xy+x+1=r 变换成一个分式,保证整除 #include<iostream> #in ...
- Helvetic Coding Contest 2018 online mirror (teams allowed, unrated)F3 - Lightsabers (hard)
题意:n个数字1-m,问取k个组成的set方案数 题解:假设某个数出现k次,那么生成函数为\(1+x+...+x^k\),那么假设第i个数出现ai次,结果就是\(\sum_{i=1}^m(1+x+.. ...
- CF 690C3. Brain Network (hard) from Helvetic Coding Contest 2016 online mirror (teams, unrated)
题目描述 Brain Network (hard) 这个问题就是给出一个不断加边的树,保证每一次加边之后都只有一个连通块(每一次连的点都是之前出现过的),问每一次加边之后树的直径. 算法 每一次增加一 ...
- [Helvetic Coding Contest 2017 online mirror]
来自FallDream的博客,未经允许,请勿转载,谢谢, 第一次在cf上打acm...和同校大佬组队打 总共15题,比较鬼畜,最后勉强过了10题. AB一样的题目,不同数据范围,一起讲吧 你有一个背包 ...
- 【Codeforces】Helvetic Coding Contest 2017 online mirror比赛记
第一次打ACM赛制的团队赛,感觉还行: 好吧主要是切水题: 开场先挑着做五道EASY,他们分给我D题,woc什么玩意,还泊松分布,我连题都读不懂好吗! 果断弃掉了,换了M和J,然后切掉了,看N题: l ...
- Helvetic Coding Contest 2016 online mirror A1
Description Tonight is brain dinner night and all zombies will gather together to scarf down some de ...
随机推荐
- Spyder的汉化
我准备写下spyder的汉化问题:对于英文大佬,从来没得汉化问题,但是对于新手和英语差的来说,汉化还是有必要,至少用汉化过得软件能快速掌握软件等.后期会用软件了在慢慢习惯英文也不迟...哈哈哈哈.本文 ...
- strsep strpbrk
#include <stdio.h> #include <string.h> int main(void) { char s[] = "aa,bb,cc.11,22, ...
- extjs4.0 treepanel节点的选中、展开! 数据的重新加载
1.extjs4.0API较3.0有非常大变化2.多级子父节点的选中和展开.3.数据的重新加载.tree.getStore().load4.节点的移除,从树中根据ID获取节点 tree.getStor ...
- 百度dureos CMake Error
CMake Error: CMAKE_C_COMPILER not set, after EnableLanguage CMake Error: CMAKE_CXX_COMPILER not set, ...
- Spark基本原理
仅作<Spark快速大数据分析>学习笔记 定义:Spark是一个用来实现 快速 而 通用 的集群计算平台:(通用的大数据处理引擎:) 改进了原Hadoop MapReduce处理模型,体现 ...
- chan
第一单元:分型.笔.线段 ?1 分型 一.分型.笔和线段所属范畴 缠师在<教你炒股票72:本ID已有课程的再梳理>中对缠论做过这样的说明“本ID的理论,本质上分两部分,一是形态学,二是动 ...
- pageHelper没有分页效果的问题
配置完全都没有问题 springboot pagehelper分页怎么都不管用 而且所有的信息记录全部都查出来了 解决方法: PageHelper.startPage(pageNum,pageSize ...
- JAVASCRIPT技术 表达式验证
失去焦点时,调用方法: <html> <head><script type="text/javascript">function upperCa ...
- hdu 1043 Eight(双向bfs)
题意:经典八数码问题 思路:双向bfs ps:还有a*算法(还不会)等解法. 代码: #include<iostream> #include<stdio.h> #include ...
- ubuntu docker的安装和使用
Docker CE for Ubuntu Docker CE for Ubuntu is the best way to install the Docker platform on Ubuntu L ...