1634: [Usaco2007 Jan]Protecting the Flowers 护花

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 448  Solved: 276
[Submit][Status]

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

题解:
同国王游戏。
代码:
 #include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<string>
#define inf 1000000000
#define maxn 100000+100
#define maxm 500+100
#define eps 1e-10
#define ll long long
#define pa pair<int,int>
using namespace std;
inline int read()
{
int x=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=*x+ch-'';ch=getchar();}
return x*f;
}
struct rec{int x,y;}a[maxn];
int n;
inline bool cmp(rec a,rec b)
{
return a.x*b.y<a.y*b.x;
}
int main()
{
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
n=read();
ll sum=;
for(int i=;i<=n;i++)a[i].x=read(),a[i].y=read(),sum+=a[i].y;
sort(a+,a+n+,cmp);
ll ans=;
for(int i=;i<=n;i++)
{
sum-=a[i].y;
ans+=sum**a[i].x;
}
printf("%lld\n",ans);
return ;
}

BZOJ1634: [Usaco2007 Jan]Protecting the Flowers 护花的更多相关文章

  1. [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 ...

  2. [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 ...

  3. BZOJ 1634: [Usaco2007 Jan]Protecting the Flowers 护花( 贪心 )

    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 ------------------------------------------- ...

  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

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

  6. 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 ...

  7. 【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× ...

  8. 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 ...

  9. 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 ...

随机推荐

  1. java GBK字符转换成为UTF-8编码字符

    import java.util.HashMap; import java.util.Map; /** * 创建日期: 2014-04-18 10:36:25 * 作者: 黄飞 * mail:huan ...

  2. HDFS集群balance(4)-- 测试计划

    转载请注明博客地址:http://blog.csdn.net/suileisl HDFS集群balance,对应版本balance design 6 如需word版本,请QQ522173163联系索要 ...

  3. [RxJS] Creation operators: fromEventPattern, fromEvent

    Besides converting arrays and promises to Observables, we can also convert other structures to Obser ...

  4. Xcode7 国际化

    1.第一步 HaiTing_xcodeproj.png 2.第二不 HaiTing_xcodeproj 2.png 3.第三步 Localizable_strings.png 5第五步 ZLBMeVi ...

  5. android自定义倒计时控件示例

    这篇文章主要介绍了Android秒杀倒计时自定义TextView示例,大家参考使用吧 自定义TextView控件TimeTextView代码: 复制代码 代码如下: import android.co ...

  6. 第一篇!in和exists性能比较和使用

    首先,先看下in和exists的区别: in 是把外表和内表作hash 连接: exists是对外表作loop循环,每次loop循环再对内表进行查询. 普遍的观点是exists比in效率高的.但是这不 ...

  7. Sandcastle Help File Builder使用教程

    Sandcastle Help File Builder相信很多的园友用过,小弟我最近因为工作原因需要生成公司的一套SDK的帮助文档,因此找了一些资料,发现网上的资料很多,但是都不怎么完全,有些只是随 ...

  8. NYOJ-569最大公约数之和

    题目链接:http://acm.nyist.net/JudgeOnline/problem.php?pid=569 此题目可以用筛选法的思想来做,但是用到一个欧拉函数 gcd(1,12)=1,gcd( ...

  9. java04 Sacnner的使用

    import java.util.Scanner; /** * 所有在java.lang包下面的所有类 不需要显示的引入包! * java.util.Scanner : 想获取用户的输入 必须引入相关 ...

  10. web04--cookie

    1.创建1.jsp <body> <form action="cookie/2.jsp" method="post"> 姓名:<i ...