hdu 1011 树形dp

Starship Troopers

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16467    Accepted Submission(s): 4396

Problem Description

You, the leader of Starship Troopers, are sent to destroy a base of the bugs. The base is built underground. It is actually a huge cavern, which consists of many rooms connected with tunnels. Each room is occupied by some bugs, and their brains hide in some of the rooms. Scientists have just developed a new weapon and want to experiment it on some brains. Your task is to destroy the whole base, and capture as many brains as possible.

To kill all the bugs is always easier than to capture their brains. A map is drawn for you, with all the rooms marked by the amount of bugs inside, and the possibility of containing a brain. The cavern‘s structure is like a tree in such a way that there is one unique path leading to each room from the entrance. To finish the battle as soon as possible, you do not want to wait for the troopers to clear a room before advancing to the next one, instead you have to leave some troopers at each room passed to fight all the bugs inside. The troopers never re-enter a room where they have visited before.

A starship trooper can fight against 20 bugs. Since you do not have enough troopers, you can only take some of the rooms and let the nerve gas do the rest of the job. At the mean time, you should maximize the possibility of capturing a brain. To simplify the problem, just maximize the sum of all the possibilities of containing brains for the taken rooms. Making such a plan is a difficult job. You need the help of a computer.

Input

The input contains several test cases. The first line of each test case contains two integers N (0 < N <= 100) and M (0 <= M <= 100), which are the number of rooms in the cavern and the number of starship troopers you have, respectively. The following N lines give the description of the rooms. Each line contains two non-negative integers -- the amount of bugs inside and the possibility of containing a brain, respectively. The next N - 1 lines give the description of tunnels. Each tunnel is described by two integers, which are the indices of the two rooms it connects. Rooms are numbered from 1 and room 1 is the entrance to the cavern.

The last test case is followed by two -1‘s.

Output

For each test case, print on a single line the maximum sum of all the possibilities of containing brains for the taken rooms.

Sample Input

5 10
50 10
40 10
40 20
65 30
70 30
1 2
1 3
2 4
2 5
1 1
20 7
-1 -1

Sample Output

50
7

Author

XU, Chuan

简单的dp,其实并不简单,我做了有一段时间。dp的题做了没几个。但是dp都是特别神奇的东西。

简单的说一下这道题的思路:

首先用vector 保存这颗树,然后用dfs记录来遍历整个树。

dp[i][j]=max(dp[i][j],dp[i][j-k]+dp[son][k].

dp[i][j]表示第i颗树用j个人能获得的最大的brain值。

时间: 2024-10-13 04:47:06

hdu 1011 树形dp的相关文章

hdu 1011 树形dp+背包

题意:有n个房间结构可看成一棵树,有m个士兵,从1号房间开始让士兵向相邻的房间出发,每个房间有一定的敌人,每个士兵可以对抗20个敌人,士兵在某个房间对抗敌人使无法走开,同时有一个价值,问你花费这m个士兵可以得到的最大价值是多少 分析:树形dp,对于点u,dp[u][j]表示以u为根的树消耗j个士兵得到的最大值,dp[i][j]=max(dp[i][j],dp[i][j-k]+dp[son][k]+val[u]) 注意是无向图,vis位置不能随便放,且注意dp不能直接+val,因为这样根节点就加不

HDU 1011 (树形DP+背包)

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=1011 题目大意:树上取点,先取父亲,再取儿子.每个点,权为w,花费为cost,给定m消费总额,求最大权和. 解题思路: 树形背包模板题.首先建一个无向图. 每个点的cost=(bug[root]+19)/20,即虫子数不满20也要派一个人. 用dp[i][j]表示以i为根的子树中,花费为j的最大权和. 转移方程:dp[i][j]=max(dp[i][j],dp[i][j-k]+dp[t][k]),

Starship Troopers(HDU 1011 树形DP)

题意: 给定n个定点和m个士兵,n个定点最终构成一棵树,每个定点有一定x个bugs和y个value,每20个bug需要消耗一个士兵,不足20也消耗一个,然后最终收获y个value,只有父节点被占领后子节点才有被占领的可能. DP状态转移方程: dp[p][j]=max(dp[p][j],dp[p][j-k]+dp[t][k]); 看的王大神的代码,DFS写的,先从叶子节点开始向上遍历进行动态规划,自己看了dp方程也没写出来.. 1 #include <iostream> 2 #include

hdu 4123 树形DP+RMQ

http://acm.hdu.edu.cn/showproblem.php?pid=4123 Problem Description Bob wants to hold a race to encourage people to do sports. He has got trouble in choosing the route. There are N houses and N - 1 roads in his village. Each road connects two houses,

hdu 1250 树形DP

Anniversary party Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Appoint description:  System Crawler  (2014-07-27) Description There is going to be a party to celebrate the 80-th Anniversary of the Ural St

hdu 4276(树形dp)

题意:带权树上有起点终点每个点上有宝藏,一个人只有T分钟要从起点到重点,问你最多能收集多少宝藏. 思路:树形dp,首先判断能不能走到终点,然后把路径上的边权变为0时间减去所有边权.dp[v][j]表示从v出发回到v话费j分钟最多能收集到的宝藏. dp[v][j] = max(dp[v][j], dp[x][k] + dp[v][j-k-2*val]); 被G++卡了好长时间,换成c++就过了. 代码如下: 1 #include <stdio.h> 2 #include <string.h

hdu 5148 树形dp+分组背包问题

http://acm.hdu.edu.cn/showproblem.php?pid=5148 Problem Description Long long ago,there is a knight called JayYe.He lives in a small country.This country is made up of n cities connected by n-1 roads(that means it's a tree).The king wants to reward Ja

HDU 1520 树形dp裸题

1.HDU 1520  Anniversary party 2.总结:第一道树形dp,有点纠结 题意:公司聚会,员工与直接上司不能同时来,求最大权值和 #include<iostream> #include<cstring> #include<cmath> #include<queue> #include<algorithm> #include<cstdio> #define max(a,b) a>b?a:b using nam

hdu 3586 树形dp+二分

题目大意:给定n个敌方据点,1为司令部,其他点各有一条边相连构成一棵 树,每条边都有一个权值cost表示破坏这条边的费用,叶子节点为前线.现要切断前线和司令部的联系,每次切断边的费用不能超过上限limit,问切断所 有前线与司令部联系所花费的总费用少于m时的最小limit.1<=n<=1000,1<=m<=100万 题目要问的是最小的最大限制,必然二分答案 然后对于每一个值,树形DP判定是否可行 dp[i]表示要切断以i为根的其它所有子树的最小代价. 其中设定叶子结点的代价为无穷大