POJ 3253 Fence Repair (哈夫曼树)
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 19660 | Accepted: 6236 |
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
3
8
5
8
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).
Source
#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<cmath>
#include<algorithm> using namespace std; int n; int main(){ //freopen("input.txt","r",stdin); priority_queue<int,vector<int>,greater<int> > q;
while(~scanf("%d",&n)){
int x;
while(!q.empty())
q.pop();
for(int i=;i<n;i++){
scanf("%d",&x);
q.push(x);
}
long long ans=; //注意精度
while(q.size()>){
int a=q.top(); q.pop();
int b=q.top(); q.pop();
int tmp=a+b;
ans+=tmp;
q.push(tmp);
}
cout<<ans<<endl;
}
return ;
}
POJ 3253 Fence Repair (哈夫曼树)的更多相关文章
- Poj 3253 Fence Repair(哈夫曼树)
Description Farmer John wants to repair a small length of the fence around the pasture. He measures ...
- poj 3253 Fence Repair (哈夫曼树 优先队列)
题目:http://poj.org/problem?id=3253 没用long long wrong 了一次 #include <iostream> #include<cstdio ...
- BZOJ 3253 Fence Repair 哈夫曼树 水题
http://poj.org/problem?id=3253 这道题约等于合并果子,但是通过这道题能够看出来哈夫曼树是什么了. #include<cstdio> #include<c ...
- POJ 3253 Fence Repair(哈夫曼编码)
题目链接:http://poj.org/problem?id=3253 题目大意: 有一个农夫要把一个木板钜成几块给定长度的小木板,每次锯都要收取一定费用,这个费用就是当前锯的这个木版的长度 给定各个 ...
- POJ 3253 Fence Repair【哈弗曼树/贪心/优先队列】
Fence Repair Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 53645 Accepted: 17670 De ...
- POJ 3253 Fence Repair(简单哈弗曼树_水过)
题目大意:原题链接 锯木板,锯木板的长度就是花费.比如你要锯成长度为8 5 8的木板,最简单的方式是把21的木板割成13,8,花费21,再把13割成5,8,花费13,共计34,当然也可以先割成16,5 ...
- POJ 3253 Fence Repair(修篱笆)
POJ 3253 Fence Repair(修篱笆) Time Limit: 2000MS Memory Limit: 65536K [Description] [题目描述] Farmer Joh ...
- poj 3253 Fence Repair 优先队列
poj 3253 Fence Repair 优先队列 Description Farmer John wants to repair a small length of the fence aroun ...
- POJ 3253 Fence Repair (优先队列)
POJ 3253 Fence Repair (优先队列) Farmer John wants to repair a small length of the fence around the past ...
- poj 3253 Fence Repair(优先队列+哈夫曼树)
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 ...
随机推荐
- 转:git设置过滤忽略的文件或文件夹
from: https://www.cnblogs.com/foohack/p/4629255.html git设置过滤忽略的文件或文件夹 我们一般向代码仓库提交项目的时候,一般需要忽略编译生成的 ...
- CodeForces 569B Inventory 货物编号
原题: http://codeforces.com/contest/569/problem/B 题目: Inventory time limit per test1 second memory lim ...
- Android Studio中关于9-patch格式图片的编译错误
最近在编译Android Studio开发的项目中在使用了9宫图后出现了编译错误,尝试了多种方法未能解决,最后仔细查看出错的日志发现,居然是图片的原因,图片中包含有alpah通道所以在执行app:me ...
- SpringMVC验证框架Validation特殊用法
基本用法不说了,网上例子很多,这里主要介绍下比较特殊情况下使用的方法. 1. 分组 有的时候,我们对一个实体类需要有多中验证方式,在不同的情况下使用不同验证方式,比如说对于一个实体类来的id来说,保存 ...
- WIN10系统 截图或者某些程序时屏幕会自动放大怎么办
右击这个应用程序,兼容性,以兼容模式运行,同时勾选高DPI设置时禁止显示缩放即可
- ZH奶酪:如何在Ubuntu上安装Java/管理多个JAVA/设置JAVA_HOME
0.简介 Java的地位及重要性,大家都懂的,很多软件都依赖于jdk,在Ubuntu上安装Java的选择有很多,openJDK,Oracle Jdk... 1.安装默认 JRE/JDK(可选) 这是最 ...
- IPC's epoch 6 is less than the last promised epoch 7
一.错误起因 Active NameNode日志出现异常IPC‘s epoch [X] is less than the last promised epoch [X+1],出现短期的双Active ...
- VMware用于Site Recovery Manager 5的vSphere Replication功能一览
http://www.searchstorage.com.cn/showcontent_54838.htm 参考:深度解析SRM 5.0和vSphere Replication http://wenk ...
- C++ 第四课:ASCII 码表
下面的 ASCII 码表包含数值在0-127之间的字符的十进制.八进制以及十六进制表示. 十进制 八进制 十六进制 字符 描述 0 0 00 NUL 1 1 01 SOH start of hea ...
- MySQL 数据类型的简单选择
选择合适的数据类型:char和varchar: +---------+------------+ | char(6) | varchar(6) | +---------+------------+ | ...