poj 3253 Fence Repair
来源:互联网 发布:传智播客java视频下载 编辑:程序博客网 时间:2024/06/15 23:28
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) units. He then purchases a single long board just long enough to saw into the N planks (i.e., whose length is the sum of the lengths Li). FJ is ignoring the "kerf", the extra length lost to sawdust when a sawcut is made; you should ignore it, too.
FJ sadly realizes that he doesn't own a saw with which to cut the wood, so he mosies over to Farmer Don's Farm with this long board and politely asks if he may borrow a saw.
Farmer Don, a closet capitalist, doesn't lend FJ a saw but instead offers to charge Farmer John for each of the N-1 cuts in the plank. The charge to cut a piece of wood is exactly equal to its length. Cutting a plank of length 21 costs 21 cents.
Farmer Don then lets Farmer John decide the order and locations to cut the plank. Help Farmer John determine the minimum amount of money he can spend to create the N planks. FJ knows that he can cut the board in various different orders which will result in different charges since the resulting intermediate planks are of different lengths.
Input
Lines 2..N+1: Each line contains a single integer describing the length of a needed plank
Output
Sample Input
3858
Sample Output
34
Hint
The original board measures 8+5+8=21. The first cut will cost 21, and should be used to cut the board into pieces measuring 13 and 8. The second cut will cost 13, and should be used to cut the 13 into 8 and 5. This would cost 21+13=34. If the 21 was cut into 16 and 5 instead, the second cut would cost 16 for a total of 37 (which is more than 34).
题意:
将一个长度为L的木板分成n个长度的小木板
每次切木板时开销为大木板的长度
比如长度为21的木板切成5,8,8,时
21->13,8 开销为21
13->5,8 开销是13
总开销为:21+13=34
求最小开销
思路:
只需每次取出最短的两块,并把两块长度之和加到集合中
直到集合中剩下一个木板为止
#include <queue>#include <cstdio>#include <cstring>#include <iostream>using namespace std;priority_queue<int,vector<int>,greater<int> > q;int main(){ long long s; int i,n,k,u,v; while(~scanf("%d",&n)){ while(!q.empty()) q.pop(); while(n--){ scanf("%d",&k); q.push(k); } s=0; while(q.size()>1){ u=q.top(); q.pop(); v=q.top(); q.pop(); s+=u+v; q.push(u+v); } cout<<s<<endl; } return 0;}
- POJ 3253 Fence Repair
- poj 3253 Fence Repair
- POJ 3253 Fence Repair
- POJ 3253 Fence Repair
- POJ 3253 Fence Repair
- POJ 3253 Fence Repair
- poj 3253 Fence Repair
- POJ 3253Fence Repair
- POJ--3253 -- Fence Repair
- poj-3253-Fence Repair
- POJ 3253 Fence Repair
- POJ 3253 Fence Repair
- poj 3253 Fence Repair
- poj 3253 Fence Repair
- POJ - 3253 Fence Repair
- POJ 3253 Fence Repair
- poj 3253 Fence Repair
- poj 3253---Fence Repair
- adb常用命令介绍
- Max Script|控制器-旋转约束_lookat
- Django入门:多对多模型
- ARM CMSIS DSP库函数arm_sin_cos_f32的BUG
- Kafka设计解析(五)- Kafka性能测试方法及Benchmark报告
- poj 3253 Fence Repair
- 欢迎使用CSDN-markdown编辑器
- 表单提交
- android小知识点
- 【DirectX11】第五篇 Hello,Shaders!
- 不错的第三方
- java.lang.NoClassDefFoundError: Could not initialize class org.hibernate.validator.internal.engine.m
- 关于TortoiseSVN的全局忽略列表
- https原理及tomcat配置https方法