poj 3253 Fence repair
来源:互联网 发布:java中属性是什么意思 编辑:程序博客网 时间:2024/06/06 08:52
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).
一个贪心问题,可以用哈夫曼树做
#include<algorithm>#include<stdio.h>using namespace std;int main(){ __int64 a[20000],n,sum=0,b,i=0,j;//这里要声明__int64,输出用%I64d,c++中的64位扩展 scanf("%I64d",&n); for(i=0;i<n;i++) { scanf("%I64d",&a[i]); } sort(a,a+n); for(i=0;i<n-1;i++)//这个循环是算法核心,每次取最小的两个结合 { b=a[i]+a[i+1]; j=i+1; while(b>a[j]&&j<n) { a[j-1]=a[j]; j++; } a[j-1]=b; sum+=b; } printf("%I64d",sum);}
- 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
- 链表版本图的深度优先和广度优先遍历
- Qt, QT/E, Qtopia 的区别
- 用例子详细介绍各种字符集编码转换问题
- C++ 中struct与class的区别
- HDU 1848 Fibonacci again and again
- poj 3253 Fence repair
- DataGridView用法笔记
- servlet配置及初始化参数
- exp导出表数据为0行
- 关于请求转发和重定向(一)--- 请求转发
- ubuntu 安装svn
- java连接Excel数据库读取,写入,操纵Excel表格
- 如何实现一个简单地Filter
- 2.4 对齐