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) 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

Line 1: One integer N, the number of planks

Lines 2..N+1: Each line contains a single integer describing the length of a needed plank

Output

Line 1: One integer: the minimum amount of money he must spend to make N-1 cuts

Sample Input

3

8

5

8

Sample Output

34

Hint

He wants to cut a board of length 21 into pieces of lengths 8, 5, and 8.

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).

优先队列实现

#include <map>
#include <list>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define eps 1e-9
#define LL long long
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define CRR fclose(stdin)
#define CWW fclose(stdout)
#define RR freopen("input.txt","r",stdin)
#pragma comment(linker, "/STACK:102400000")
#define WW freopen("output.txt","w",stdout) const int MAX = 6000000+5; int main()
{
priority_queue<LL,vector<LL>, greater<LL> >Q;
int n;
LL sum,a;
while(~scanf("%d",&n))
{
while(n--)
{
scanf("%lld",&a);
Q.push(a);
}
sum=0;
while(!Q.empty())
{
int u = Q.top();
Q.pop();
if(!Q.empty())
{
int v=Q.top();
Q.pop();
sum+=(u+v);
Q.push(u+v);
}
}
printf("%lld\n",sum);
}
return 0;
}

版权声明:本文为博主原创文章,未经博主允许不得转载。

哈夫曼树-Fence Repair 分类: 树 POJ 2015-08-05 21:25 2人阅读 评论(0) 收藏的更多相关文章

  1. A Plug for UNIX 分类: POJ 图论 函数 2015-08-10 14:18 2人阅读 评论(0) 收藏

    A Plug for UNIX Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 14786 Accepted: 4994 Desc ...

  2. Hiking 分类: 比赛 HDU 函数 2015-08-09 21:24 3人阅读 评论(0) 收藏

    Hiking Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others) Total Subm ...

  3. 哈希-4 Values whose Sum is 0 分类: POJ 哈希 2015-08-07 09:51 3人阅读 评论(0) 收藏

    4 Values whose Sum is 0 Time Limit: 15000MS Memory Limit: 228000K Total Submissions: 17875 Accepted: ...

  4. 哈希-Gold Balanced Lineup 分类: POJ 哈希 2015-08-07 09:04 2人阅读 评论(0) 收藏

    Gold Balanced Lineup Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 13215 Accepted: 3873 ...

  5. 哈希-Snowflake Snow Snowflakes 分类: POJ 哈希 2015-08-06 20:53 2人阅读 评论(0) 收藏

    Snowflake Snow Snowflakes Time Limit: 4000MS Memory Limit: 65536K Total Submissions: 34762 Accepted: ...

  6. android开发之调试技巧 分类: android 学习笔记 2015-07-18 21:30 140人阅读 评论(0) 收藏

    我们都知道,android的调试打了断点之后运行时要使用debug as->android application 但是这样的运行效率非常低,那么我们有没有快速的方法呢? 当然有. 我们打完断点 ...

  7. UI基础:UIButton.UIimage 分类: iOS学习-UI 2015-07-01 21:39 85人阅读 评论(0) 收藏

    UIButton是ios中用来响应用户点击事件的控件.继承自UIControl 1.创建控件 UIButton *button=[UIButton buttonWithType:UIButtonTyp ...

  8. UI基础:UITextField 分类: iOS学习-UI 2015-07-01 21:07 68人阅读 评论(0) 收藏

    UITextField 继承自UIControl,他是在UILabel基础上,对了文本的编辑.可以允许用户输入和编辑文本 UITextField的使用步骤 1.创建控件 UITextField *te ...

  9. 架构师速成6.7-设计开发思路-uml 分类: 架构师速成 2015-07-29 18:25 157人阅读 评论(0) 收藏

    uml是什么东西?统一建模语言,一门语言,是用来进行软件设计的一门语言. 其实一门语言的诞生并不伟大,让大多数人都使用才足够伟大.uml就是一门伟大的语言,因为目前软件设计的唯一语言就是它. UML其 ...

随机推荐

  1. 3n+1问题

    猜想: 对于任意大于1的自然数n,若n为奇数,则将n变为3n+1,否则变为n的一半. 经过若干次这样的变换,一定会使n变为1.例如3->10->5->16->8->2-& ...

  2. Leetcode: Create Maximum Number

    Given two arrays of length m and n with digits 0-9 representing two numbers. Create the maximum numb ...

  3. SQL查询一个表的总记录数的方法

    一.简单查询语句 1. 查看表结构 SQL>DESC emp; 2. 查询所有列 SQL>SELECT * FROM emp; 3. 查询指定列 SQL>SELECT empmo, ...

  4. table创建固定表头

    布局:两个div,上部内容将表头复制,高度固定,下部div内部将table设置为margin:-**px; 隐藏掉表头,下部div设置overflow,即可. 代码:

  5. C++之路进阶——codevs2404(糖果)

    2404 糖果 2011年省队选拔赛四川  时间限制: 1 s  空间限制: 128000 KB  题目等级 : 大师 Master     题目描述 Description 幼儿园里有N个小朋友,l ...

  6. 分享Centos作为WEB服务器的防火墙规则

    # Firewall configuration written by system-config-firewall # Manual customization of this file is no ...

  7. android中在代码中设置margin属性

    1,不多说,小知识点,直接上代码 LinearLayout.LayoutParams layoutParams = new LinearLayout.LayoutParams(15, 15);// 创 ...

  8. c语言小程序

    这是一个用c语言写的小程序,功能是随机输出30道100以内的四则运算,先生成两个随机数,再通过随机数确定四则运算符号,最后输出题目. #include<iostream> using na ...

  9. BJFU 1009

    描述 现在社会上的抽奖活动简直是太多了.前段时间中国联通就举办了一个很无聊的抽奖活动,规则是每人可以向中国联通的短信系统发送一个实数,系统每天会从这些数字中选择一个无重复(就是有且只有一个)且最小的数 ...

  10. ORACLE 默认密码确认

    select USER_NAME USER_WITH_DEFAULT_PASSWORD from ( select fnd_web_sec.validate_login('AME_INVALID_AP ...