题意 有n个洞穴编号为1-n,洞穴间有通道,形成了一个n-1条边的树, 洞穴的入口即根节点是1. 每个洞穴有x只bugs,并有价值y的金子,全部消灭完一个洞穴的虫子,就可以获得这个洞穴的y个金子. 现在要派m个战士去找金子,从入口进入.每次只有消灭完当前洞穴的所有虫子,才可以选择进入下一个洞穴. 一个战士可以消灭20只虫子,如果要杀死x只虫子,那么要x/20向上取整即(x+19)/20个战士. 如果要获得某个洞穴的金子,必须留下足够杀死所有虫子的战士数量, 即(x+19)/20个战士,然后这些留…

Starship Troopers Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 8540    Accepted Submission(s): 2379 Problem Description You, the leader of Starship Troopers, are sent to destroy a base of t…

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…

Starship Troopers Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 17498    Accepted Submission(s): 4644 Problem Description You, the leader of Starship Troopers, are sent to destroy a base of t…

Starship Troopers Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) 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 h…

Starship Troopers Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 19362 Accepted Submission(s): 5130 Problem Description You, the leader of Starship Troopers, are sent to destroy a base of the bu…

本来是一道很水的树形DP题 设dp[i][j]表示,带着j个人去攻打以节点i为根的子树的最大收益 结果wa了一整晚 原因: 坑点1: 即使这个节点里面没有守卫,你如果想获得这个节点的收益,你还是必须派一个人去这个节点,不然谁帮你去拿收益? 坑点2: 题目说是从节点1开始攻打,然后我就以为给出的数据都是以1为根节点的,就没有加双向边 不过, 最后我加了双向边,然后还是wa了 又找了很久,最后发现是前向星的数组开小了 注意:数组开小了可能会返回wa,tle,re等等等等 #include<cstdi…

Starship Troopers Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 20833    Accepted Submission(s): 5550 Problem Description   You, the leader of Starship Troopers, are sent to destroy a base of…

Starship Troopers Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 11476    Accepted Submission(s): 3166  Problem Description You, the leader of Starship Troopers, are sent to destroy a base of…

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, an…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1011 dp[u][i]为以u为根节点的,花了不超过i元钱能够得到的最大价值 因为题目里说要访问子节点必须先访问根节点,所以 dp[u][i+j] = max(dp[u][i],dp[u][i]+dp[v][j]); v是u的儿子节点. 代码: import java.util.*; import static java.lang.Math.*; class Graph{ ArrayList<Arra…

http://acm.hdu.edu.cn/showproblem.php?pid=1011   题意:每个节点有两个值bug和brain,当清扫该节点的所有bug时就得到brain值,只有当父节点被清空时,才可以清扫它          的子节点,而清扫需要一定的人员.给定M个人员,N个结点的树,求最大brain和   这看起来是一道非常简单的背包dp 但是 写完提交wa之后,我发现这道题,并不简单!因为他的题意并不是完全和我找到的题意一样(我承认我语文很差劲所以随便找个题解看题意..),题目…

题意是说在一个洞穴中有许多房间,每个房间中有一些虫子和大脑,这些房间之间用隧道相连形成一棵树,士兵们杀虫子的能力有限,也可以直接杀死虫子而不消耗士兵战斗力,但这样就无法得到房间中的大脑,士兵们不能走回头路,问给定士兵数量时能得到的大脑最大值. 在树上进行动态规划,对于每个节点来说,选择了它,就要损失士兵战斗力,不选择它,就可以将这些省下来的士兵战斗力用在后面的房间中,后面的房间将最优选择传递到当前位置,以此来判断从而得到最优解.转移方程:dp[ i ][ j ] = max(dp[ i ][ j…

本文出自   http://blog.csdn.net/shuangde800 题目链接 : hdu-1011   题意 有n个洞穴编号为1-n,洞穴间有通道,形成了一个n-1条边的树, 洞穴的入口即根节点是1. 每个洞穴有x只bugs,并有价值y的金子,全部消灭完一个洞穴的虫子,就可以获得这个洞穴的y个金子. 现在要派m个战士去找金子,从入口进入.每次只有消灭完当前洞穴的所有虫子,才可以选择进入下一个洞穴. 一个战士可以消灭20只虫子,如果要杀死x只虫子,那么要x/20向上取整即(x+19)/…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1011 题意: 题目大意是有n个房间组成一棵树,你有m个士兵,从1号房间开始让士兵向相邻的房间出发,每个房间里有一个代价,代价是值/20个士兵, 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. 思路: 树上背包,这题比较坑爹.士兵为0,输出0.要是一个房间的cost不足20的倍数也要补全20的倍数. dp[i][j]表示以i节点为子树的root使用j个士兵的最大价值( 不用管父节点 ),…

Starship Troopers 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 occupie…

Anniversary party Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 12770    Accepted Submission(s): 5142 Problem Description There is going to be a party to celebrate the 80-th Anniversary of the…

Starship Troopers Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 15457    Accepted Submission(s): 4152 Problem Description You, the leader of Starship Troopers, are sent to destroy a base of t…

http://acm.split.hdu.edu.cn/showproblem.php?pid=3586 题意: 给定一个带权无向树,要切断所有叶子节点和1号节点(总根)的联系,每次切断边的费用不能超过上限limit,问在保证总费用<=m下的最小的limit. 思路: 对于上限limit我们可以二分查找.然后就是树形dp,看代码就可以理解的. #include<iostream> #include<algorithm> #include<cstring> #inc…

zxa and leaf 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5682 Description zxa have an unrooted tree with n nodes, including (n−1) undirected edges, whose nodes are numbered from 1 to n. The degree of each node is defined as the number of the edge…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6201 题意:给出一棵树,每个点有一个权值,代表商品的售价,树上每一条边上也有一个权值,代表从这条边经过所需要的花费.现在需要你在树上选择两个点,一个作为买入商品的点,一个作为卖出商品的点,当然需要考虑从买入点到卖出点经过边的花费.使得收益最大.允许买入点和卖出点重合,即收益最小值为0. 解法:我们设1为根节点,假设一开始一个人身上的钱为0.我们设dp[i][0]表示从根节点走到i及其子树并中任一点买…

Warm up Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others) Total Submission(s): 3160    Accepted Submission(s): 718 Problem Description N planets are connected by M bidirectional channels that allow instant transport…

The more, The Better Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4859    Accepted Submission(s): 2864 Problem Description ACboy很喜欢玩一种战略游戏,在一个地图上,有N座城堡,每座城堡都有一定的宝物,在每次游戏中ACboy允许攻克M个城堡并获得里面的宝…

题目链接:http://acm.hust.edu.cn/vjudge/contest/view.action?cid=106048#problem/B 题意:给出一些点相连,找出最小的点数覆盖所有的 用树形dp来解决 dp[i][0]表示i点不选,dp[i][1]表示i点选,dp[i][0] = sum{ dp[ son[i] ][1] }, // i不选,则i的子节点全选, dp[i][1] = sum { min( dp[ son[i] ][0], dp[ son[i] ][1] ) } +…

第一道树形DP.很容易理解. #include <cstdio> #include <cstring> #include <cstdlib> #define MAXN 105 typedef struct { int n, p; } room_t; room_t rooms[MAXN]; int adj[MAXN][MAXN]; int dp[MAXN][MAXN]; bool visit[MAXN]; int n, m; int max(int a, int b) {…

CGZ大佬提醒我,我要是再不更博客可就连一月一更的频率也没有了... emmm,正好做了一道有点意思的题,就拿出来充数吧=.= 题意 一棵树,有 $ n (n\leq50000) $ 个节点,每个点都有一个颜色,共有 $ k(k\leq10) $ 种颜色,问有多少条路径可以遍历到所有 $ k $ 种颜色?(一条路径交换起点终点就算两条哦) 做法 事实证明,连我都能不看题解想出来的题果然都是水题qwq 我是从CJ的xzyxzy大佬上的博客上看到这道题的,所以就理所当然用FWT做了...然后才发现网…

transaction transaction transaction Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 132768/132768 K (Java/Others)Total Submission(s): 895    Accepted Submission(s): 441 Problem Description Kelukin is a businessman. Every day, he travels aroun…

Anniversary party Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 8667    Accepted Submission(s): 3748 Problem Description There is going to be a party to celebrate the 80-th Anniversary of the…

题目链接 题意:给你一棵树(必须从根节点出发),每个节点上都有bug和value,你有m个骑士,每个骑士能消灭20个bug,你必须消灭该节点的全部bug才能拿到该节点的value,问最多能拿到value是多少. 思路:典型的背包dp ,  dp[n][m]=max(dp[n][m-x]+value,dp[n][m]) #include<iostream> #include<cstdio> using namespace std; #define N 110 #define INF…

HDU 1561  The more, The Better 题目大意就不说了 直接DP[i][j]表示i为跟节点的子树上攻克j个城堡的所能获得的最多宝物的数量 DP[fa][j] = MAX{DP[fa][i-k] + DP[child][k]}; 首先一个问题就是说如果子树u下的任意子节点被选择了,那么u是一定需要选择的,怎么在DP时保证准确性,其实这个很好解决,我们在计算时是需要枚举k(子节点的攻克数量)的,那么我们迫使k<j就可以了,这样的话DP[fa][j] 就不会被子节点的DP[ch…