Find Metal Mineral Time Limit: 1000ms Memory Limit: 65768KB This problem will be judged on HDU. Original ID: 400364-bit integer IO format: %I64d      Java class name: Main     Humans have discovered a kind of new metal mineral on Mars which are distr…

Find Metal Mineral Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)Total Submission(s): 3397    Accepted Submission(s): 1588 Problem Description Humans have discovered a kind of new metal mineral on Mars which are d…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4003 Humans have discovered a kind of new metal mineral on Mars which are distributed in point‐like with paths connecting each of them which formed a tree. Now Humans launches k robots on Mars to collect…

Find Metal Mineral Problem Description Humans have discovered a kind of new metal mineral on Mars which are distributed in point‐like with paths connecting each of them which formed a tree. Now Humans launches k robots on Mars to collect them, and du…

看别人思路的 树形分组背包. 题意:给出结点数n,起点s,机器人数k,然后n-1行给出相互连接的两个点,还有这条路线的价值,要求最小花费 思路:这是我从别人博客里找到的解释,因为很详细就引用了 dp[i][j]表示对于以i结点为根结点的子树,放j个机器人所需要的权值和. 当j=0时表示放了一个机器人下去,遍历完结点后又回到i结点了.状态转移方程类似背包 如果最终的状态中以i为根结点的树中有j(j>0)个机器人,那么不可能有别的机器人r到了这棵树后又跑到别的树中去 因为那样的话,一定会比j中的某一…

Find Metal Mineral Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others) Total Submission(s): 2018    Accepted Submission(s): 913 Problem Description Humans have discovered a kind of new metal mineral on Mars which are d…

[HDU4003]Find Metal Mineral Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)Total Submission(s): 3686    Accepted Submission(s): 1723 Problem Description Humans have discovered a kind of new metal mineral on Mars wh…

Find Metal Mineral Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others) Total Submission(s): 2371    Accepted Submission(s): 1079 Problem Description Humans  have  discovered  a  kind  of  new  metal  mineral  on  Mars …

题意: 一棵有权树,从根结点中放入 K 个机器人.求用这 K 个机器人遍历全部的结点最少的权值和. 思路: 1. dp[u][i] 表示给以 u 为根节点的子树放 i 个机器人,遍历其子树所须要的最小权值. 2. 关键在于 dp[u][0] 的理解,表示:最后停留在以 u 为根节点的子树下 0 个机器人,而且遍历了 u 子树的最小权值和. 3. 以下的步骤就变成和分组背包类似的情况了,根节点 u 给孩子 v 放多少个机器人. 4. dp[u][i] = min(dp[u][i], dp[u][j…

题目链接 很棒的一个树形DP.学的太渣了. #include <cstdio> #include <string> #include <cstring> #include <iostream> #include <algorithm> using namespace std; ][]; struct node { int u,v,w,next; }edge[]; ],t,flag[]; int n,m; void CL() { t = ; mem…