题目描述 农夫约翰为了修理栅栏,要将一块很长的木板切割成N块.准备切成的木板长度为L1,L2,L3--LN,未切割前木板的长度恰好为切割后木板长度的总和.每次切断木板时,需要的开销为这块木板的长度.请求出按照目标要求将木板切割完的最小开销是多少?例如长度为21的木板切割成长度为13和8,开销为21:把长度为13的木板切割成5和8,则开销为13,所以将长度为21的木板切割成8,5,8的三块,开销是34. 样例输入 N = 3, L = {8,5,8} 样例输出 34 思路分析 以 3 5 8 5为…

Farmer John wants to repair a small length of the fence around the pasture. He measures the fence and finds that he needs N (1 ≤ N ≤ 20,000) planks of wood, each having some integer length Li (1 ≤ Li ≤ 50,000) units. He then purchases a single long b…

Farmer John wants to repair a small length of the fence around the pasture. He measures the fence and finds that he needs N (1 ≤ N ≤ 20,000) planks of wood, each having some integer length Li (1 ≤ Li ≤ 50,000) units. He then purchases a single long b…

农夫约翰为了修理栅栏,要将一块很长的木板切割成N块.准备切成的木板长度为L1.L2.L3...LN,未切割前的木板长度恰好为切割后木板长度的总和.每次切断木板时,需要的开销为这块木板的长度.例如长度为21的木板要切成长度为5,8,8的三块木板.长度为21的木板切成长度为13和8的板时,开销为21.再将长度为5和8的板时,开销为13.于是合计开销是34.请求按照目标要求将木板切割完最小的开销是多少. #include "iostream" using namespace std; typ…

题意:给出一段无限长的棍子,切一刀需要的代价是棍子的总长,例如21切一刀 变成什么长度 都是代价21 列如7切成5 和2 也是代价7题解:可以利用霍夫曼编码的思想 短的棍子就放在底层 长的尽量切少一次 直接用优先队列 取前2个和一个然后代价加起来就好 有一个小trick就是 只有一个棍子的时候要特判 还有sum要开Long long #include<vector> #include<cstdio> #include<iostream> #include<queu…

Fence Repair Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status Practice POJ 3253 Appoint description:  hanjiangtao  (2014-11-12) System Crawler  (2015-04-24) Description Farmer John wants to repair a small len…

Fence Repair Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 77001   Accepted: 25185 Description Farmer John wants to repair a small length of the fence around the pasture. He measures the fence and finds that he needs N (1 ≤ N ≤ 20,000)…

Fence Repair Farmer John wants to repair a small length of the fence around the pasture. He measures the fence and finds that he needs N (1 ≤ N ≤ 20,000) planks of wood, each having some integer length Li (1 ≤ Li ≤ 50,000) units. He then purchases a…

地址 http://poj.org/problem?id=3253 题解 本题是<挑战程序设计>一书的例题 根据树中描述 所有切割的代价 可以形成一颗二叉树 而最后的代价总和是与子节点和深度相关的 由于切割的次数是确定的 该二叉树的节点就是确定的. 也就是说我们可以贪心的处理  最小长度的子节点放在最下面 如图 ac代码如下 使用了堆 记录每次最小的元素 堆的使用真的不是很方便 , 另外还需要注意 爆int 所以需要使用long long 记录元素的和 #include <iostrea…

Fence Repair Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 53645   Accepted: 17670 Description Farmer John wants to repair a small length of the fence around the pasture. He measures the fence and finds that he needs N (1 ≤ N ≤ 20,000)…