考虑相邻的两头奶牛 a , b , 我们发现它们顺序交换并不会影响到其他的 , 所以我们可以直接按照这个进行排序

---------------------------------------------------------------------------------

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
 
#define rep( i , n ) for( int i = 0 ; i < n ; ++i )
#define clr( x , c ) memset( x , c , sizeof( x ) )
 
using namespace std;
 
typedef long long ll;
 
const int maxn = 100000 + 5;
 
struct data {
int t , w;
void Read() {
scanf( "%d%d" , &t , &w );
t <<= 1;
}
bool operator < ( const data &rhs ) const {
return t * rhs.w < w * rhs.t;
}
};
 
data A[ maxn ];
 
int main() {
freopen( "test.in" , "r" , stdin );
int n;
cin >> n;
rep( i , n )
   A[ i ].Read();
   
sort( A , A + n );
ll ans = 0;
int cnt = 0;
for( int i = n - 1 ; i >= 0 ; i-- ) {
ans += 1LL * cnt * A[ i ].t;
cnt += A[ i ].w;
}
cout << ans << "\n";
return 0;
}

---------------------------------------------------------------------------------

1634: [Usaco2007 Jan]Protecting the Flowers 护花

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 515  Solved: 329
[Submit][Status][Discuss]

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 cows were in his garden eating his beautiful flowers. Wanting to minimize the subsequent damage, FJ decided to take immediate action and transport the cows back to their barn. Each cow i is at a location that is Ti minutes (1 <= Ti <= 2,000,000) away from the 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 the barn. Moving cow i to the barn requires 2*Ti minutes (Ti to get there and Ti to return). 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.

   约翰留下他的N只奶牛上山采木.他离开的时候,她们像往常一样悠闲地在草场里吃草.可是,当他回来的时候,他看到了一幕惨剧:牛们正躲在他的花园里,啃食着他心爱的美丽花朵!为了使接下来花朵的损失最小,约翰赶紧采取行动,把牛们送回牛棚. 牛们从1到N编号.第i只牛所在的位置距离牛棚Ti(1≤Ti《2000000)分钟的路程,而在约翰开始送她回牛棚之前,她每分钟会啃食Di(1≤Di≤100)朵鲜花.无论多么努力,约翰一次只能送一只牛回棚.而运送第第i只牛事实上需要2Ti分钟,因为来回都需要时间.    写一个程序来决定约翰运送奶牛的顺序,使最终被吞食的花朵数量最小.

Input

* Line 1: A single integer

N * Lines 2..N+1: Each line contains two space-separated integers, Ti and Di, that describe a single cow's characteristics

第1行输入N,之后N行每行输入两个整数Ti和Di.

Output

* Line 1: A single integer that is the minimum number of destroyed flowers

一个整数,表示最小数量的花朵被吞食.

Sample Input

6
3 1
2 5
2 3
3 2
4 1
1 6

Sample Output

86

HINT

约翰用6,2,3,4,1,5的顺序来运送他的奶牛.

Source

BZOJ 1634: [Usaco2007 Jan]Protecting the Flowers 护花( 贪心 )的更多相关文章

  1. BZOJ 1634: [Usaco2007 Jan]Protecting the Flowers 护花

    Description Farmer John went to cut some wood and left N (2 <= N <= 100,000) cows eating the g ...

  2. BZOJ 1634 [Usaco2007 Jan]Protecting the Flowers 护花:贪心【局部分析法】

    题目链接:http://www.lydsy.com/JudgeOnline/problem.php?id=1634 题意: 约翰留下他的N只奶牛上山采木.可是,当他回来的时候,他看到了一幕惨剧:牛们正 ...

  3. bzoj 1634: [Usaco2007 Jan]Protecting the Flowers 护花【贪心】

    因为交换相邻两头牛对其他牛没有影响,所以可以通过交换相邻两头来使答案变小.按照a.t*b.f排降序,模拟着计算答案 #include<iostream> #include<cstdi ...

  4. 1634: [Usaco2007 Jan]Protecting the Flowers 护花

    1634: [Usaco2007 Jan]Protecting the Flowers 护花 Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 493  So ...

  5. [BZOJ1634][Usaco2007 Jan]Protecting the Flowers 护花 贪心

    1634: [Usaco2007 Jan]Protecting the Flowers 护花 Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 885  So ...

  6. 【BZOJ】1634: [Usaco2007 Jan]Protecting the Flowers 护花(贪心)

    http://www.lydsy.com/JudgeOnline/problem.php?id=1634 贪心.. 我们发现,两个相邻的牛(a和b)哪个先走对其它的牛无影响,但是可以通过 a的破坏花× ...

  7. 【bzoj1634】[Usaco2007 Jan]Protecting the Flowers 护花 贪心

    题目描述 Farmer John went to cut some wood and left N (2 <= N <= 100,000) cows eating the grass, a ...

  8. BZOJ1634: [Usaco2007 Jan]Protecting the Flowers 护花

    1634: [Usaco2007 Jan]Protecting the Flowers 护花 Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 448  So ...

  9. [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

  1. 在输出视图中使用使用html注释会导致在Chrome中css失效

    在做SportsStore例子时,在视图List.cshtml中使用了html注释,即 <!-- 注释 --> 结果在加载css时,chrome浏览器中所有css效果都失效.IE不受影响. ...

  2. 5_Navigation Bar

    5 // // ViewController.swift // Navigation Bar // // Created by ZC on 16/1/9. // Copyright © 2016年 Z ...

  3. OpenCV初探

    一种基于OpenCV的PHP图像人脸识别技术 openCV是一个开源的用C/C++开发的计算机图形图像库,非常强大,研究资料很齐全.本文重点是介绍如何使用php来调用其中的局部的功能.人脸侦查技术只是 ...

  4. 时尚B2B方兴未艾-Maker’s Row 获100万美元种子投资 |华丽志

    时尚B2B方兴未艾-Maker's Row 获100万美元种子投资 |华丽志 华丽志 » 网internet, 时尚B2B方兴未艾-Maker's Row 获100万美元种子投资 由 luxeco 发 ...

  5. 《Effective C++》:条款48:理解力template 元编程

    Template metaprogramming(TMP,模板元编程)这是写template-based C++规划.编译过程.template metaprogramming随着C++写模板程序,化 ...

  6. NSFileManager的应用

    单例,是在 一个文件中只创建一次就能够全部一起共享,多创建的地址是相同的 NSFileManager *manager=[NSFileManager defaultManager]; 是一个单例   ...

  7. visual studio 2013 使用IIS Express附加调试MVC5

    1.如何找到调试的站点的进程[由于图片无法上传,就不上传图片了] 2.vs运行的时候,在状态栏会存在一个IIS Express 进程,点击显示所有的应用程序,找到想要调试的程序的PID; 3.附加调试 ...

  8. jackson的简单实用实例(json)

    一个json格式的字符串比如: {"status":10001,"code":"HDkGzI","pubkey":&qu ...

  9. Java "==和equals区别"

    本文转载至中软国际电子政务部Jeff Chi总结,转载请说明出处. 概述:        A.==可用于基本类型和引用类型:当用于基本类型时候,是比较值是否相同:当用于引用类型的时候,是比较对象是否相 ...

  10. Mongodb PHP开发类库

    <?php /** * Mongodb 基本操作API,支持基本类似关系统型数据库的操作接口 * * @version 1.0 * * [说明] * * 1:该版本API实现了 Mongodb ...