Fence Repair Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 33114   Accepted: 10693 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)…

题目大意:背景大概是个资本家剥削工人剩余价值的故事....有一块木板,要把它切成几个长度,切一次的费用是这整块被切木板的长度,例如将一个长度为21的木板切成2和19两块费用为21,切成两块的长度及顺序是可以自己定的,问最小费用是多少 思路:一个很明显的贪心思路是每次将最长切下来,这样后续切割就不会用到这根最长的木板,结果也就是最优的了.具体操作的时候可以倒过来想:有一系列切完的小木板,要将它们合并成一块大木板,费用为合并完后那个大木板的长度,问最小费用.这样,每次合并最小木板的思路就跃然纸上了-…

POJ 3253 Fence Repair(修篱笆) Time Limit: 2000MS   Memory Limit: 65536K [Description] [题目描述] 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, eac…

POJ 3253 Fence Repair (优先队列) Farmer John wants to repair a small length of the fence around the pasture. He measures the fence and finds that he needsN (1 ≤ N ≤ 20,000) planks of wood, each having some integer lengthLi (1 ≤ Li ≤ 50,000) units. He the…

poj 3253 Fence Repair 优先队列 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) planks of wood, each having some integer length Li (1 ≤ Li ≤ 50,000) u…

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: 23913   Accepted: 7595 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 Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 32424 Accepted: 10417 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) pla…

题目地址:POJ 3253 哈夫曼树的结构就是一个二叉树,每个父节点都是两个子节点的和. 这个题就是能够从子节点向根节点推. 每次选择两个最小的进行合并.将合并后的值继续加进优先队列中.直至还剩下一个元素为止. 代码例如以下: #include <iostream> #include <stdio.h> #include <string.h> #include <stdlib.h> #include <math.h> #include <c…

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)…