题目

In a village called Byteville, there are houses connected with N-1 roads. For each pair of houses, there is a unique way to get from one to another. The houses are numbered from 1 to . The house no. 1 belongs to the village administrator Byteasar. As part of enabling modern technologies for rural areas framework, computers have been delivered to Byteasar's house. Every house is to be supplied with a computer, and it is Byteasar's task to distribute them. The citizens of Byteville have already agreed to play the most recent version of FarmCraft (the game) as soon as they have their computers. Byteasar has loaded all the computers on his pickup truck and is about to set out to deliver the goods. He has just the right amount of gasoline to drive each road twice. In each house, Byteasar leaves one computer, and immediately continues on his route. In each house, as soon as house dwellers get their computer, they turn it on and install FarmCraft. The time it takes to install and set up the game very much depends on one's tech savviness, which is fortunately known for each household. After he delivers all the computers, Byteasar will come back to his house and install the game on his computer. The travel time along each road linking two houses is exactly 1 minute, and (due to citizens' eagerness to play) the time to unload a computer is negligible. Help Byteasar in determining a delivery order that allows all Byteville's citizens (including Byteasar) to start playing together as soon as possible. In other words, find an order that minimizes the time when everyone has FarmCraft installed.

mhy 住在一棵有n个点的树的1号结点上,每个结点上都有一个妹子。 mhy从自己家出发,去给每一个妹子都送一台电脑,每个妹子拿到电脑后就会开始安装zhx牌杀毒软件,第i个妹子安装时间为Ci。 树上的每条边mhy能且仅能走两次,每次耗费1单位时间。mhy送完所有电脑后会回自己家里然后开始装zhx牌杀毒软件。 卸货和装电脑是不需要时间的。 求所有妹子和mhy都装好zhx牌杀毒软件的最短时间。

Input

The first line of the standard input contains a single integer N(2<=N<=5 00 000) that gives the number of houses in Byteville. The second line contains N integers C1,C2…Cn(1<=Ci<=10^9), separated by single spaces; Ci is the installation time (in minutes) for the dwellers of house no. i. 
The next N-1 lines specify the roads linking the houses. Each such line contains two positive integers a and b(1<=a<b<=N) , separated by a single space. These indicate that there is a direct road between the houses no. a and b.

Output

The first and only line of the standard output should contain a single integer: the (minimum) number of minutes after which all citizens will be able to play FarmCraft together.

Sample Input

6 1 8 9 6 3 2 1 3 2 3 3 4 4 5 4 6

Sample Output

11

HINT

Explanation: Byteasar should deliver the computers to the houses in the following order: 3, 2, 4, 5, 6, and 1. The game will be installed after 11, 10, 10, 10, 8, and 9 minutes respectively, in the house number order. Thus everyone can play after 11 minutes. If Byteasar delivered the game in the following order: 3, 4, 5, 6, 2, and 1, then the game would be installed after: 11, 16, 10, 8, 6, and 7 minutes respectively. Hence, everyone could play only after 16 minutes,

题解:

设f[x]为x的子树中全部安完软件的时间,对于x节点的a,b儿子,设他们的子树大小为siz[a],siz[b],显然遍历一遍子树a的时间就是2*siz[a],然后分类讨论

若先走a,后走b,那么f[x]=max(f[a]+1,f[b]+2siz[a]+1) 若先走b,后走a,那么f[x]=max(f[a]+2siz[b]+1,f[b]+1)

对于f[a]+1和f[b]+1我们可以不用考虑,那么如果先走a更优,当且仅当:

f[b]+2siz[a]+1<f[a]+2siz[b]+1

转移

f[a]-2siz[a]>f[b]-2siz[b]

那么直接将x的儿子按照f[i]-2siz[i]排序,先走f[i]-2siz[i]的就好了

Attention!!:根节点的软件是最后安装的,所以要特判,ans=max(2*(n-1)+1+time[1],f[1])

代码

(有些丑不要介意)

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn=;
int n,cnt;
int to[maxn<<],next[maxn<<],head[maxn],v[maxn],p[maxn],f[maxn],siz[maxn];
void add(int a,int b){
to[cnt]=b,next[cnt]=head[a],head[a]=cnt++;
}
bool cmp(int a,int b){
return f[a]-*siz[a]>f[b]-*siz[b];
}
void dfs(int x,int fa){
int i,sum=;
siz[x]=;
for(i=head[x];i!=-;i=next[i])
if(to[i]!=fa)
dfs(to[i],x),siz[x]+=siz[to[i]];
p[]=;
for(i=head[x];i!=-;i=next[i])
if(to[i]!=fa)
p[++p[]]=to[i];
sort(p+,p+p[]+,cmp);
if(x!=)f[x]=v[x];
for(i=;i<=p[];i++)
f[x]=max(f[x],f[p[i]]+sum+),sum+=*siz[p[i]];
}
int main(){
scanf("%d",&n);
int i,a,b;
memset(head,-,sizeof(head));
for(i=;i<=n;i++) scanf("%d",&v[i]);
for(i=;i<n;i++) scanf("%d%d",&a,&b),add(a,b),add(b,a);
dfs(,);
printf("%d",max(f[],*(siz[]-)+v[]));
return ;
}

【树形dp】Farmcraft的更多相关文章

  1. [POI2014][树形DP]FarmCraft

    题目 In a village called Byteville, there are houses connected with N-1 roads. For each pair of houses ...

  2. bzoj 3829: [Poi2014]FarmCraft 树形dp+贪心

    题意: $mhy$ 住在一棵有 $n$ 个点的树的 $1$ 号结点上,每个结点上都有一个妹子. $mhy$ 从自己家出发,去给每一个妹子都送一台电脑,每个妹子拿到电脑后就会开始安装 $zhx$ 牌杀毒 ...

  3. poj3417 LCA + 树形dp

    Network Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 4478   Accepted: 1292 Descripti ...

  4. COGS 2532. [HZOI 2016]树之美 树形dp

    可以发现这道题的数据范围有些奇怪,为毛n辣么大,而k只有10 我们从树形dp的角度来考虑这个问题. 如果我们设f[x][k]表示与x距离为k的点的数量,那么我们可以O(1)回答一个询问 可是这样的话d ...

  5. 【BZOJ-4726】Sabota? 树形DP

    4726: [POI2017]Sabota? Time Limit: 20 Sec  Memory Limit: 128 MBSec  Special JudgeSubmit: 128  Solved ...

  6. 树形DP+DFS序+树状数组 HDOJ 5293 Tree chain problem(树链问题)

    题目链接 题意: 有n个点的一棵树.其中树上有m条已知的链,每条链有一个权值.从中选出任意个不相交的链使得链的权值和最大. 思路: 树形DP.设dp[i]表示i的子树下的最优权值和,sum[i]表示不 ...

  7. 树形DP

    切题ing!!!!! HDU  2196 Anniversary party 经典树形DP,以前写的太搓了,终于学会简单写法了.... #include <iostream> #inclu ...

  8. BZOJ 2286 消耗战 (虚树+树形DP)

    给出一个n节点的无向树,每条边都有一个边权,给出m个询问,每个询问询问ki个点,问切掉一些边后使得这些顶点无法与顶点1连接.最少的边权和是多少.(n<=250000,sigma(ki)<= ...

  9. POJ2342 树形dp

    原题:http://poj.org/problem?id=2342 树形dp入门题. 我们让dp[i][0]表示第i个人不去,dp[i][1]表示第i个人去 ,根据题意我们可以很容易的得到如下递推公式 ...

  10. hdu1561 The more, The Better (树形dp+背包)

    题目链接:http://acm.split.hdu.edu.cn/showproblem.php?pid=1561 思路:树形dp+01背包 //看注释可以懂 用vector建树更简单. 代码: #i ...

随机推荐

  1. Java实现 LeetCode 748 最短完整词(字母拆分+暴力)

    748. 最短完整词 如果单词列表(words)中的一个单词包含牌照(licensePlate)中所有的字母,那么我们称之为完整词.在所有完整词中,最短的单词我们称之为最短完整词. 单词在匹配牌照中的 ...

  2. Java实现 LeetCode 698 划分为k个相等的子集(递归)

    698. 划分为k个相等的子集 给定一个整数数组 nums 和一个正整数 k,找出是否有可能把这个数组分成 k 个非空子集,其总和都相等. 示例 1: 输入: nums = [4, 3, 2, 3, ...

  3. Java实现 LeetCode 617 合并二叉树(遍历树)

    617. 合并二叉树 给定两个二叉树,想象当你将它们中的一个覆盖到另一个上时,两个二叉树的一些节点便会重叠. 你需要将他们合并为一个新的二叉树.合并的规则是如果两个节点重叠,那么将他们的值相加作为节点 ...

  4. Java实现 LeetCode 373 查找和最小的K对数字

    373. 查找和最小的K对数字 给定两个以升序排列的整形数组 nums1 和 nums2, 以及一个整数 k. 定义一对值 (u,v),其中第一个元素来自 nums1,第二个元素来自 nums2. 找 ...

  5. Java实现 LeetCode 279 完全平方数

    279. 完全平方数 给定正整数 n,找到若干个完全平方数(比如 1, 4, 9, 16, -)使得它们的和等于 n.你需要让组成和的完全平方数的个数最少. 示例 1: 输入: n = 12 输出: ...

  6. Java实现 蓝桥杯VIP 算法提高 士兵排队问题

    算法提高 士兵排队问题 时间限制:1.0s 内存限制:256.0MB 试题 有N个士兵(1≤N≤26),编号依次为A,B,C,-,队列训练时,指挥官要把一些士兵从高到矮一次排成一行,但现在指挥官不能直 ...

  7. Linux 自动挂载与fstab文件修复

    /etc/fstab文件 自动挂载就是写入/etc/fstab文件 vi /etc/fstab 其中,第九行是/分区的自动挂载信息,有6个字段 第一字段表示分区的UUID(硬盘通用唯一识别码,使用du ...

  8. vue+jquery使用FormData向后端传递数据和文件,express如何获取

    使用multiparty 模块 下载 cnpm install multiparty --save 前端代码: <template> <div class="add-are ...

  9. How to delete a directory recursively in Java

    在java8或更高版本中,使用NIO API递归删除一个非空目录: try { // 创建stream流 Stream<Path> file = Files.walk(Paths.get( ...

  10. iOS-AutoLayout中动画使用的细节 和 iOS layout机制

    在Main.storyboard拖入一个UIView,随便设置一个背景色, 使用autolayout  为紫色的view添加约束 :(0,0,100,100) , 为该view添加动画代码如下: #i ...