1018. Binary Apple Tree

Time limit: 1.0 second
Memory limit: 64 MB
Let's imagine how apple tree looks in binary computer world. You're right, it looks just like a binary tree, i.e. any biparous branch splits up to exactly two new branches. We will enumerate by integers the root of binary apple tree, points of branching and the ends of twigs. This way we may distinguish different branches by their ending points. We will assume that root of tree always is numbered by 1 and all numbers used for enumerating are numbered in range from 1 to N, where N is the total number of all enumerated points. For instance in the picture below N is equal to 5. Here is an example of an enumerated tree with four branches:
2   5
\ /
3 4
\ /
1
As you may know it's not convenient to pick an apples from a tree when there are too much of branches. That's why some of them should be removed from a tree. But you are interested in removing branches in the way of minimal loss of apples. So your are given amounts of apples on a branches and amount of branches that should be preserved. Your task is to determine how many apples can remain on a tree after removing of excessive branches.

Input

First line of input contains two numbers: N and Q (2 ≤ N ≤ 100; 1 ≤ Q ≤ N − 1). N denotes the number of enumerated points in a tree. Q denotes amount of branches that should be preserved. NextN − 1 lines contains descriptions of branches. Each description consists of a three integer numbers divided by spaces. The first two of them define branch by it's ending points. The third number defines the number of apples on this branch. You may assume that no branch contains more than 30000 apples.

Output

Output should contain the only number — amount of apples that can be preserved. And don't forget to preserve tree's root ;-)

Sample

input output
5 2
1 3 1
1 4 10
2 3 20
3 5 20
21

题目大意:n个点 n-1条边,现在要保留Q条边,求保留下的边的去权值和的最大值。

把边的权值映射到点上,边的权值相当于这个点指向根节点的权值,所以问题转换成对点的操作。

先统计出以当前点为根节点的子树的点数(包括当前根节点),然后dp,

这里dp,以u为根节点保留j个点能得到最大值,状态转移方程

dp[u][j]=max(dp[u][j],dp[u][j-k]+dp[v]k]+val)

val是v到u的权值。

/* ***********************************************
Author :guanjun
Created Time :2016/10/15 15:43:48
File Name :timus1018.cpp
************************************************ */
#include <bits/stdc++.h>
#define ull unsigned long long
#define ll long long
#define mod 90001
#define INF 0x3f3f3f3f
#define maxn 10010
#define cle(a) memset(a,0,sizeof(a))
const ull inf = 1LL << ;
const double eps=1e-;
using namespace std;
priority_queue<int,vector<int>,greater<int> >pq;
struct Node{
int x,y;
};
struct cmp{
bool operator()(Node a,Node b){
if(a.x==b.x) return a.y> b.y;
return a.x>b.x;
}
}; bool cmp(int a,int b){
return a>b;
}
struct node{
int y;
int val;
}; vector<node>v[];
int sz[],n,m,num;
int dp[][];
void dfs(int u,int fa){
num++;
sz[u]=;
for(int i=;i<v[u].size();i++){
int y=v[u][i].y;
if(y==fa)continue;
dfs(y,u);
sz[u]+=sz[y];
}
}
void dfs2(int u,int fa){
for(int i=;i<v[u].size();i++){
int y=v[u][i].y;
int val=v[u][i].val;
if(y==fa)continue;
//cout<<u<<" "<<sz[u]<<endl;
dfs2(y,u);
for(int j=sz[u];j>;j--){
for(int k=;k<j;k++){
dp[u][j]=max(dp[u][j],dp[u][j-k]+dp[y][k]+val);
}
}
}
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
#endif
//freopen("out.txt","w",stdout);
while(cin>>n>>m){
num=;
int x,y,z;
cle(sz);
for(int i=;i<n;i++){
cin>>x>>y>>z;
v[x].push_back({y,z});
v[y].push_back({x,z});
}
cle(dp);
dfs(,-);
dfs2(,-);
cout<<dp[][m+]<<endl;
}
return ;
}

timus 1018. Binary Apple Tree的更多相关文章

  1. CJOJ 1976 二叉苹果树 / URAL 1018 Binary Apple Tree(树型动态规划)

    CJOJ 1976 二叉苹果树 / URAL 1018 Binary Apple Tree(树型动态规划) Description 有一棵苹果树,如果树枝有分叉,一定是分2叉(就是说没有只有1个儿子的 ...

  2. URAL 1018 Binary Apple Tree(树DP)

    Let's imagine how apple tree looks in binary computer world. You're right, it looks just like a bina ...

  3. Ural 1018 Binary Apple Tree

    题目链接:http://acm.timus.ru/problem.aspx?space=1&num=1018 Dynamic Programming. 首先要根据input建立树形结构,然后在 ...

  4. ural 1018 Binary Apple Tree(树形dp | 经典)

    本文出自   http://blog.csdn.net/shuangde800 ------------------------------------------------------------ ...

  5. BNUOJ 13358 Binary Apple Tree

    Binary Apple Tree Time Limit: 1000ms Memory Limit: 16384KB This problem will be judged on Ural. Orig ...

  6. 【URAL 1018】Binary Apple Tree

    http://vjudge.net/problem/17662 loli蜜汁(面向高一)树形dp水题 #include<cstdio> #include<cstring> #i ...

  7. URAL1018 Binary Apple Tree(树形DP)

    题目大概说一棵n结点二叉苹果树,n-1个分支,每个分支各有苹果,1是根,要删掉若干个分支,保留q个分支,问最多能保留几个苹果. 挺简单的树形DP,因为是二叉树,都不需要树上背包什么的. dp[u][k ...

  8. URAL1018 Binary Apple Tree(树dp)

    组队赛的时候的一道题,那个时候想了一下感觉dp不怎么好写呀,现在写了出来,交上去过了,但是我觉得我还是应该WA的呀,因为总感觉dp的不对. #pragma warning(disable:4996) ...

  9. URAL1018. Binary Apple Tree

    链接 简单树形DP #include <iostream> #include<cstdio> #include<cstring> #include<algor ...

随机推荐

  1. Linux系统的启动流程

    Linux系统的启动流程: 1.通电(通常按下电源键,开始通电) 2.加载BIOS (通常看到显示器提示按F2进入主板) 3.读取MBR (MBR硬盘的入口地址,用来装载引导) 4.进入引导 (通常有 ...

  2. Flask框架 之路由

    一.视图函数路由规则 from flask import Flask, redirect, url_for # 创建flask应用对象 # __name__ 代表当前模块名称 # flask以当前目录 ...

  3. Vue指令4:v-on

    监听事件 事件:click\keydown <button v-on:click="greet"></button> 可以简写为  <button @ ...

  4. 100 道 Linux 笔试题,能拿 80 分就算大神!

    本套笔试题共100题,每题1分,共100分.(参考答案在文章末尾) 1. cron 后台常驻程序 (daemon) 用于: A. 负责文件在网络中的共享 B. 管理打印子系统C. 跟踪管理系统信息和错 ...

  5. c# Dictionary 扩展方法

    主要用于接口请求,数据转换 #region Dictionary 扩展方法 public static string getString(this Dictionary<string, stri ...

  6. [Git]Please make sure you have the correct access rights and the repository exists

    这个问题是这样,需要在已有github账号的A机器上,再创建一个github账号,新账号创建完毕,将代码通过机器A push上之后,再另一台机器B,clone 这个项目时报出了如下错误: Permis ...

  7. ORACLE 查看当前用户信息(用户,表视图,索引,表空间,同义词,存储过程,约束条件)

    1.用户 查看当前用户的缺省表空间 SQL>select username,default_tablespace from user_users; 查看当前用户的角色 SQL>select ...

  8. Java核心技术 卷一 复习笔记(乙

    1.字符串从概念上讲,Java字符串就是Unicode字符序列.Java没有内置的字符串类型,而是在标准Java类库中提供了一个预定义类,叫String. 每个用双引号括起来的字符串都是 String ...

  9. Discuz 部署,500 – 内部服务器错误。 您查找的资源存在问题,因而无法显示。

    Windows7 IIS 500 – 内部服务器错误解决方案 1.解决方法:打开IIS,在功能视图中找到“错误页”,双击进去后,看最右边的“操作”下的“编辑功能设置…”,将“错误响应”下的“详细错误” ...

  10. Entity SQL rules for Wrapped and Unwrapped Results

    Here are some rules to remember for Entity SQL queries: 1.Use SELECT VALUE when projecting more than ...