倒过来看 , 每次总是选择最短的两块木板合并 , 用heap维护 ------------------------------------------------------------------------------ #include<cstdio> #include<cstring> #include<algorithm> #include<iostream> #include<queue>   #define rep( i , n )…

Description Farmer John想修理牧场栅栏的某些小段.为此,他需要N(1<=N<=20,000)块特定长度的木板,第i块木板的长度为Li(1<=Li<=50,000).然后,FJ去买了一块很长的木板,它的长度正好等于所有需要的木板的长度和.接下来的工作,当然是把它锯成需要的长度.FJ忽略所有切割时的损失——你也应当忽略它. FJ郁闷地发现,他并没有锯子来把这块长木板锯开.于是他把这块长木板带到了Farmer Don的农场,想向FD借用锯子. 作为一个有商业头脑的资…

题目 1724: [Usaco2006 Nov]Fence Repair 切割木板 Time Limit: 5 Sec  Memory Limit: 64 MB Description Farmer John想修理牧场栅栏的某些小段.为此,他需要N(1<=N<=20,000)块特定长度的木板,第i块木板的长度为Li(1<=Li<=50,000).然后,FJ去买了一块很长的木板,它的长度正好等于所有需要的木板的长度和.接下来的工作,当然是把它锯成需要的长度.FJ忽略所有切割时的损失—…

1724: [Usaco2006 Nov]Fence Repair 切割木板 Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 854  Solved: 426[Submit][Status] Description Farmer John想修理牧场栅栏的某些小段.为此,他需要N(1<=N<=20,000)块特定长度的木板,第i块木板的长度为Li(1<=Li<=50,000).然后,FJ去买了一块很长的木板,它的长度正好等于所有需要的木板的…

题目链接:http://www.lydsy.com/JudgeOnline/problem.php?id=1724 题意: 你要将一块长木板切成n段,长度分别为a[i](长木板的长度 = ∑ a[i]). 每一次切割的花费为被切割木板的长度. 问你切完的最小花费. 题解: 合并果子. 反过来想:切割 = 合并 贪心策略:每次选目前所有堆中最小的两个合并.(尽可能少移动大的果子堆) 实现:优先队列 AC Code: #include <iostream> #include <stdio.h…

一开始被题目读错题= =以为每次只能割一块,那么就是从大到小切 但是其实是可以分为几堆来切的 所以可以逆着来,变为合并n个木板代价最小 易证每次找最小的两堆合并代价最小 用优先队列维护堆..偷偷懒= = #include<stdio.h> #include<string.h> #include<algorithm> #include<queue> using namespace std; priority_queue<int,vector<int…

[算法]贪心+堆 #include<cstdio> #include<algorithm> using namespace std; ; int n,heap[maxn],sz; void heap_push(int x) { heap[++sz]=x;//新数入堆底 int now=sz;//以堆底为起点 &&heap[now]<heap[now>>])//非根节点的父亲>儿子时------注意非根判断 { swap(heap[now],h…

如果反着看,看成合并木板,就和合并果子一样了,把若干块放进一个小根堆,然后每次取出两个合并,把合并结果加进答案和堆里 代码里小根堆用优先队列实现(懒 #include<iostream> #include<cstdio> #include<queue> using namespace std; const int N=20005; int n; priority_queue<int,vector<int>,greater<int> >…

Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 1356  Solved: 714[Submit][Status][Discuss] Description Farmer John想修理牧场栅栏的某些小段.为此,他需要N(1<=N<=20,000)块特定长度的木板,第i块木板的长度为Li(1<=Li<=50,000).然后,FJ去买了一块很长的木板,它的长度正好等于所有需要的木板的长度和.接下来的工作,当然是把它锯成需要的长度.FJ忽略所…

堆对于stl priority_queue ,我们自己定义的类自己重载<,对于非自定义类我们默认大根堆,如若改成小根堆则写成std::priority<int,vector<int>,greator<int> >.时间复杂度除了pop push是O(log)外都是O(1).当然手打会比stl快不少,下面介绍手打堆.对于手打堆他出来用于优先队列之外还能用于堆排序,就先建堆,然后依次取出.除已有操作以外,还有一个建堆过程,一般用于堆排序,就是一次把许多数的建成堆,就是…

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <queue> using namespace std; int n,a; typedef long long ll; ll ans; void read(int &x){ x=; ; char ch; ; +ch-'; x*=f; } priority_queue&…

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…

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…

题意:将一块木板切成N块,长度分别为:a1,a2,……an,每次切割木板的开销为当前木板的长度.求出按照要求将木板切割完毕后的最小开销. 思路:比较奇特的贪心 每次切割都会将当前木板一分为二,可以按切割要求画出二叉树. 总开销 = 各叶子节点的值 x 该叶子节点的深度 树的深度一定,为了使总开销尽可能的小,那么比较深的叶子节点的值应尽可能的小,所以二叉树应为: 模拟出所建立的二叉树,求出根节点的值即可 为了节省时间,可以利用优先队列每次取出最小的两个值合并,并将两数之和重新加入数组,直到数组内的…

Fence Repair Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 28359   Accepted: 9213 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…

题目大意: 输入N ( 1 ≤ N ≤ 20,000 ) :将一块木板分为n块 每次切割木板的开销为这块木板的长度,即将长度为21的木板分为13和8,则开销为21 接下来n行描述每块木板要求的长度Li ( 1 ≤ Li ≤ 50,000 ) 木板长度恰好等于各段木板长之和 Sample Input 3858 Sample Output 34 挑战上的一个奇妙的贪心的方法 #include <bits/stdc++.h> #define ll long long using namespace…

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)…

倒着的石子合并,注意不是取当前最长木板贪心做,而是取当前最小累加答案: 例如 4 5 6 7 若按第一种思路:ans=22+15+9 第二种:ans=22+13+9,可以先从中间某一块分开,这样答案更优 #include<bits/stdc++.h> #define rep(i,x,y) for(register int i=x;i<=y;i++) #define ll long long using namespace std; ; inline int read(){ ,x=; ;…

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)…

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…

Fence Repair 问题大意:农夫约翰为了修理栅栏,要将一块很长的木块切割成N块,准备切成的木板的长度为L1,L2...LN,未切割前的木板的长度恰好为切割后木板的长度的总和,每次切断木板的时候,需要的开销为这块木板的长度,例如长度为21的木板需要切成长度为5,8,8的三块木板,长为21的木板切成长为13和8的板时,开销为21,.再将长度为13的板切成长度为5和8的板时,开销是13,于是合计开销是34,然后要你求切割完的最小开销是什么. 乍眼看这一道题好像有很多种切割方案,似乎是很难解决的…

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)…

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…

题目链接:http://poj.org/problem?id=3253 Fence Repair Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 61005   Accepted: 20119 Description Farmer John wants to repair a small length of the fence around the pasture. He measures the fence and fi…

篱笆修理(Fence Repair) 代码(C) 本文地址: http://blog.csdn.net/caroline_wendy 题目: 把一块木板切成N块, 每次切两块, 分割的开销是木板长度, 求将木板分割完的最小开销. 即霍夫曼编码(Huffman). 贪心算法, 相似二叉树型结构, 最短板和次短板是兄弟结点, 选取两个最小木板, 最后进行分割, 合并两个最小木板, 依次递推. 代码: /* * main.cpp * * Created on: 2014.7.17 * Author:…

Fence Repair Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 42979   Accepted: 13999 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)…

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…

Fence Repair Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 51411   Accepted: 16879 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: 26167   Accepted: 8459 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)…