Game

Time Limit:1500MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Appoint description: 
System Crawler  (2015-05-26)

Description

It is well known that Keima Katsuragi is The Capturing God because of his exceptional skills and experience in ''capturing'' virtual girls in gal games. He is able to play $k$ games simultaneously.

One day he gets a new gal game named ''XX island''. There are $n$ scenes in that game, and one scene will be transformed to different scenes by choosing different options while playing the game. All the scenes form a structure like a rooted tree such that the root is exactly the opening scene while leaves are all the ending scenes. Each scene has a value , and we use $w_i$ as the value of the $i$-th scene. Once Katsuragi entering some new scene, he will get the value of that scene. However, even if Katsuragi enters some scenes for more than once, he will get $w_i$ for only once.

For his outstanding ability in playing gal games, Katsuragi is able to play the game $k$ times simultaneously. Now you are asked to calculate the maximum total value he will get by playing that game for $k$ times.

 

Input

The first line contains an integer $T$($T \le 20$), denoting the number of test cases.

For each test case, the first line contains two numbers $n, k(1 \le k \le n \le 100000)$, denoting the total number of scenes and the maximum times for Katsuragi to play the game ''XX island''.

The second line contains $n$ non-negative numbers, separated by space. The $i$-th number denotes the value of the $i$-th scene. It is guaranteed that all the values are less than or equal to $2^{31} - 1$.

In the following $n - 1$ lines, each line contains two integers $a, b(1 \le a, b \le n)$, implying we can transform from the $a$-th scene to the $b$-th scene.

We assume the first scene(i.e., the scene with index one) to be the opening scene(i.e., the root of the tree).

 

Output

For each test case, output ''Case #t:'' to represent the $t$-th case, and then output the maximum total value Katsuragi will get. 
 

Sample Input

2
5 2
4 3 2 1 1
1 2
1 5
2 3
2 4
5 3
4 3 2 1 1
1 2
1 5
2 3
2 4
 

Sample Output

Case #1: 10
Case #2: 11
 
 
题目大意:给你一棵树,有n个结点,有n-1条边,每个结点都有一个价值,可以从跟到叶子k次,但是只能拿一次价值,问你从根1到达叶子的最大价值。
 
解题思路:用深搜划分出与根临接的顶点个数数量的子树,回溯的时候从这些子树中可以找出重链,如果在中间有不是重链上的链,则把该链放入优先队列中,否则维护出一条重链。dp[i]表示从叶子到i结点的最大价值和。
 
 
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
#define LL __int64
const int maxn=1e5+100;
vector<LL>V[maxn];
priority_queue<LL>Q;
LL a[maxn];
LL dp[maxn];
LL dfs(int cur){
int i,v;
LL tmp;
dp[cur]=a[cur];
for(i=0;i<V[cur].size();i++){
v=V[cur][i];
tmp=dfs(v);
if(dp[cur]>tmp+a[cur]){ //如果有其他轻链,把轻链放入队列
Q.push(tmp);
}else{ //如果当前这个链不是最重的链,把从叶子到该结点的这段链放入队列
Q.push(dp[cur]-a[cur]);
dp[cur]=tmp+a[cur];
}
}
return dp[cur];
}
int main(){
int t ,n,m,i,j,k,u,v,cnt=0;
LL ans;
scanf("%d",&t);
while(t--){
scanf("%d%d",&n,&k);
for(i=1;i<=n;i++)
scanf("%lld",&a[i]);
for(i=1;i<n;i++){
scanf("%d%d",&u,&v);
V[u].push_back(v);
}
dfs(1);
ans=dp[1];
for(i=1;i<k;i++){
ans+=Q.top();
Q.pop();
}
printf("Case #%d: %lld\n",++cnt,ans);
while(!Q.empty())
Q.pop();
for(i=0;i<=n;i++){
V[i].clear();
}
}
return 0;
}

  

hdu 5242——Game——————【树链剖分思想】的更多相关文章

  1. HDU 5242 利用树链剖分思想进行贪心

    题目大意: 在给定带权值节点的树上从1开始不回头走到某个底端点后得到所有经过的点的权值后,这些点权值修改为0,到达底部后重新回到1,继续走,问走k次,最多能得到多少权值之和 这其实就是相当于每一次都走 ...

  2. hdu 5458 Stability(树链剖分+并查集)

    Stability Time Limit: 3000/2000 MS (Java/Others)    Memory Limit: 65535/102400 K (Java/Others)Total ...

  3. HDU 5242 树链剖分思想的贪心

    题意及博客 树链剖分分为2步,第一次求出深度,重儿子,第二次求出重链,用到了启发式的思想,即对于比较重的儿子,尽量去完整的维护它.类似于我们去合并两个堆,明显把小的堆逐个插入大的堆中会比大的往小的插更 ...

  4. HDU 5044 (树链剖分+树状数组+点/边改查)

    题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5044 题目大意:修改链上点,修改链上的边.查询所有点,查询所有边. 解题思路: 2014上海网赛的变 ...

  5. HDU 3966(树链剖分+点修改+点查询)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3966 题目大意:营地的分布成树型.每个营地都有一些人,每次修改修改一条链上的所有营地的人数,每次查询单 ...

  6. HDU 5458 Stability (树链剖分+并查集+set)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5458 给你n个点,m条边,q个操作,操作1是删边,操作2是问u到v之间的割边有多少条. 这题要倒着做才 ...

  7. HDU 5044 Tree --树链剖分

    题意:给一棵树,两种操作: ADD1: 给u-v路径上所有点加上值k, ADD2:给u-v路径上所有边加上k,初始值都为0,问最后每个点和每条边的值,输出. 解法:树链剖分可做,剖出来如果直接用线段树 ...

  8. HDU - 3966-Aragorn' Story(树链剖分+线段树)

    链接:https://vjudge.net/problem/HDU-3966 题意: Our protagonist is the handsome human prince Aragorn come ...

  9. HDU 3966 RE 树链剖分 线段树 Aragorn's Story

    题意: 给出一棵树,每个顶点上有一个权值. 操作:选择一条路径,并将路径上所有的点的权值同时加或减某个数. 查询:某个点的当前权值 分析: 树链剖分完毕后,就是简单的线段树区间更新. 提交的时候注意要 ...

  10. Tree HDU - 6547 (树链剖分,线段树)

    wls 有三棵树,树上每个节点都有一个值 ai,现在有 2 种操作: 将一条链上的所有节点的值开根号向下取整: 求一条链上值的和: 链的定义是两点之间的最短路. Input 第一行两个数 n, q 分 ...

随机推荐

  1. 《C#多线程编程实战》2.4 SemaphoreSlim

    这个简单多了. 理解也是很好理解. 比上一个mutex好理解多了. 这个SemaphoreSlim是干什么呢? 就是限制线程的来访问. 好比说一次只有两个,一次只有三个  这样的线程来访问资源. 有点 ...

  2. [SinGuLaRiTy] 2017 百度之星程序设计大赛 初赛A

    [SinGuLaRiTy-1036] Copyright (c) SinGuLaRiTy 2017. All Rights Reserved. 小C的倍数问题 Time Limit: 2000/100 ...

  3. 题解 CF500D 【New Year Santa Network】

    题目链接 这道题首先是要看看该如何化简,先把三元组化成二元组. 之后统计经过某条边的 次数$*$权值  的和. 最后除以总基数 $tot$ 其中,每条边被计算的次数为 子树的点数$*$非子树的点数 ( ...

  4. HDU6318-2018ACM暑假多校联合训练2-1010-Swaps and Inversions-树状数组

    本题题意是,给你一个长度为n的序列,使用最少的操作把序列转换为从小到大的顺序,并输出操作数*min(x,y) 实质上是算出该序列中有多少逆序对,有归并排序和树状数组两种算法,由于数据之间的差值有点大, ...

  5. TX2 Clone

    由于给TX2配置了很多的开发环境,也修改了一些驱动,想将这些环境能够完整的迁移到一块bare TX2,于是尝试了clone的方法. 这种方法的优点是: 确保了移植的TX2 与已经配置好的环境是一致的: ...

  6. vector<vector<int>> 使用简单示例

    #include <iostream> #include <vector> using namespace std; int main() { vector<vector ...

  7. undefined reference to symbol ‘_ZN2cv6String10deallocateEv

    使用qt编译Caffe时出现如下错误: undefined reference to symbol '_ZN2cv6String10deallocateEv' error adding symbols ...

  8. Mysql Update更新错误 Error Code:1175

    Mysql 5.7,默认执行 update 语句时遇到错误提示: Error Code: 1175. You are using safe update mode and you tried to u ...

  9. ASP.NET中MD5的加密方式很简单

    在ASP.NET中MD5的加密方式很简单,代码如下: FormsAuthentication.HashPasswordForStoringInConfigFile(str, "MD5&quo ...

  10. Codeforces - 185A 简单矩阵快速幂

    题意:求第n个三角形内部的上三角形个数 对每个三角形分别维护上下三角形个数,记为\(dp[1][i],dp[2][i]\) 规律很明显是 \(dp[1][i+1]=3*dp[1][i]+dp[2][i ...