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
- 在UTF-8中,一个汉字为什么需要三个字节?(转)
http://www.cnblogs.com/web21/p/6092414.html UNICODE是万能编码,包含了所有符号的编码,它规定了所有符号在计算机底层的二进制的表示顺序.有关Unicod ...
- Linux命令行极简教程
1.命令行真的好吗 程序员的使命 维基百科的解释: 命令行界面(英语:command-line interface,缩写:CLI)是在图形用户界面得到普及之前使用最为广泛的用户界面,它通常不支持鼠标, ...
- SqlServer驱动包 Maven
SqlServer驱动包 Maven 学习了:https://blog.csdn.net/wu843820873/article/details/50484623 mvn install: mvn i ...
- pdf阅读器改背景色
- ubuntu16.04下部署tomcat9和java8启动一次需要七八分钟
一.环境如下 Ubuntu16.04 +tomcat9+openjdk1.8 二.问题 在tomcat的bin下执行./startup.sh 如下图没有问题 root@bogon:/usr/apac ...
- DOM元素尺寸offsetWidth,scrollWidth,clientWidth等具体解释
样例: <div id="div" style="height: 200px;width: 200px;border:solid 50px red;overflow ...
- 迅为4412开发板Linux驱动教程——编写简单应用调用驱动
Linux驱动教程:http://pan.baidu.com/s/1c0hljUS 编写简单应用调用驱动--头文件 • 打印头文件 – include <stdio.h>调用打印函数pri ...
- .NET 开发套装
Dapper,轻量级ORM GitHub - StackExchange/dapper-dot-net: DapperSharpZipLib,ZIP压缩库 SharpZipLib by icsharp ...
- 通过项目逐步深入了解Mybatis<二>
Mybatis 解决 jdbc 编程的问题 1. 数据库链接创建.释放频繁造成系统资源浪费从而影响系统性能,如果使用数据库链接池可解决此问题. 解决:在SqlMapConfig.xml中配置数据链接池 ...
- HDU1069(还是dp基础)
今天不想说太多废话-由于等下要写自己主动提交机. 不知道能不能成功呢? 题目的意思就是,一个猴子,在叠砖头 ...以下的要严格大于上面的.求叠起来最高能到多少- n非常少,n^2算法毫无压力-话说dp ...