POJ 3262 Protecting the Flowers 贪心(性价比)
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 7812 | Accepted: 3151 |
Description
Farmer John went to cut some wood and left N (2 ≤ N ≤ 100,000) cows eating the grass, as usual. When he returned, he found to his horror that the cluster of cows was in his garden eating his beautiful flowers. Wanting to minimize the subsequent damage, FJ decided to take immediate action and transport each cow back to its own barn.
Each cow i is at a location that is Ti minutes (1 ≤ Ti ≤ 2,000,000) away from its own barn. Furthermore, while waiting for transport, she destroys Di (1 ≤ Di ≤ 100) flowers per minute. No matter how hard he tries, FJ can only transport one cow at a time back to her barn. Moving cow i to its barn requires 2 × Ti minutes (Ti to get there and Ti to return). FJ starts at the flower patch, transports the cow to its barn, and then walks back to the flowers, taking no extra time to get to the next cow that needs transport.
Write a program to determine the order in which FJ should pick up the cows so that the total number of flowers destroyed is minimized.
Input
Lines 2..N+1: Each line contains two space-separated integers, Ti and Di, that describe a single cow's characteristics
Output
Sample Input
6
3 1
2 5
2 3
3 2
4 1
1 6
Sample Output
86
Hint
Source
#include<stdio.h>
#include<algorithm>
using namespace std; struct Node{
double t,d;
}node[];
bool cmp(Node a,Node b)
{
return (a.t/a.d)<(b.t/b.d); //贪心
}
int main()
{
int n,i;
long long sum,ans;
scanf("%d",&n);
sum=;
for(i=;i<=n;i++){
scanf("%lf%lf",&node[i].t,&node[i].d);
sum+=node[i].d;
}
sort(node+,node+n+,cmp);
ans=;
for(i=;i<=n;i++){
sum-=node[i].d;
ans+=sum*node[i].t*;
}
printf("%lld\n",ans);
return ;
}
POJ 3262 Protecting the Flowers 贪心(性价比)的更多相关文章
- poj 3262 Protecting the Flowers 贪心 牛吃花
Protecting the Flowers Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 11402 Accepted ...
- poj -3262 Protecting the Flowers (贪心)
http://poj.org/problem?id=3262 开始一直是理解错题意了!!导致不停wa. 这题是农夫有n头牛在花园里啃花朵,然后农夫要把它们赶回棚子,每次只能赶一头牛,并且给出赶回每头牛 ...
- poj 3262 Protecting the Flowers 贪心
题意:给定n个奶牛,FJ把奶牛i从其位置送回牛棚并回到草坪要花费2*t[i]时间,同时留在草地上的奶牛j每分钟会消耗d[j]个草 求把所有奶牛送回牛棚内,所消耗草的最小值 思路:贪心,假设奶牛a和奶牛 ...
- poj 3262 Protecting the Flowers
http://poj.org/problem?id=3262 Protecting the Flowers Time Limit: 2000MS Memory Limit: 65536K Tota ...
- POJ 3262 Protecting the Flowers 【贪心】
题意:有n个牛在FJ的花园乱吃.所以FJ要赶他们回牛棚.每个牛在被赶走之前每秒吃Di个花朵.赶它回去FJ来回要花的总时间是Ti×2.在被赶走的过程中,被赶走的牛就不能乱吃 思路: 先赶走破坏力大的牛假 ...
- POJ 3362 Protecting the Flowers
这题和金华区域赛A题(HDU 4442)是一样的做法. 对两个奶牛进行分析,选择两个奶牛总花费少的方式排序. bool cmp(const X&a,const X&b){ return ...
- 【POJ - 3262】Protecting the Flowers(贪心)
Protecting the Flowers 直接中文 Descriptions FJ去砍树,然后和平时一样留了 N (2 ≤ N ≤ 100,000)头牛吃草.当他回来的时候,他发现奶牛们正在津津有 ...
- POJ3262 Protecting the Flowers 【贪心】
Protecting the Flowers Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 4418 Accepted: ...
- [bzoj1634][Usaco2007 Jan]Protecting the Flowers 护花_贪心
Protecting the Flowers 护花 bzoj-1634 Usaco-2007 Jan 题目大意:n头牛,每头牛有两个参数t和atk.表示弄走这头牛需要2*t秒,这头牛每秒会啃食atk朵 ...
随机推荐
- Active Directory虚拟机搭建域控服务器环境
前言 还是和上一章一样,痛苦过后还是记录下给后来人提供便利为妙. 虚拟机选择:建议Hyper-V或者VMware 系统选择:建议WIindows Server 2003及以上 我这里是使用VMware ...
- PHP魔术方法之__call与__callStatic方法
<?php class human{ private function t(){ } //魔术方法__call /* $method 获得方法名 $arg 获得方法的参数集合 */ public ...
- 自己定义ActionBar标题与菜单中的文字样式
自己定义标题文字样式 标题样式是ActionBar样式的一部分,所以要先定义ActionBar的样式 <style name="AppTheme" parent=" ...
- HDU 5336 XYZ and Drops 2015 Multi-University Training Contest 4 1010
这题的题意是给你一幅图,图里面有水滴.每一个水滴都有质量,然后再给你一个起点,他会在一開始的时候向四周发射4个小水滴,假设小水滴撞上水滴,那么他们会融合,假设质量大于4了,那么就会爆炸,向四周射出质量 ...
- CAS 单点登录原理
访问服务: 浏览器发送请求访问应用系统 定向认证: 应用系统重定向用户请求到 SSO 服务器. 用户认证:用户身份认证. 发放票据: 认证通过后,SSO 服务器会产生一个随机的 Service Tic ...
- Darwin Streaming Server性能测试报告
为了验证Darwin Streaming Server在流媒体点播上的性能,EasyDarwin开源项目官方特地与国内某大型视频网站进行了一次性能测试(千兆网络环境下),针对本次RTSP直播流媒体测试 ...
- File syncing and sharing software with file encryption and group sharing, emphasis on reliability and high performance.
http://seafile.com/ showdoc haiwen/seafile: File syncing and sharing software with file encryption a ...
- delphi android 录像(使用了JMediaRecorder,MediaRecorder的使用方法)
delphi xe系列自带的控件都无法保存录像,经网友帮忙,昨天终于实现了录像功能(但有个问题是录像时无画面显示),程序主要使用了JMediaRecorder,MediaRecorder的使用方法可参 ...
- [haoi2008]玩具命名
某人有一套玩具,并想法给玩具命名.首先他选择WING四个字母中的任意一个字母作为玩具的基本名字.然后他会根据自己的喜好,将名字中任意一个字母用“WING”中任意两个字母代替,使得自己的名字能够扩充得很 ...
- Codeforces Round #173 (Div. 2) E. Sausage Maximization —— 字典树 + 前缀和
题目链接:http://codeforces.com/problemset/problem/282/E E. Sausage Maximization time limit per test 2 se ...