Assignments

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2038    Accepted Submission(s): 1013

Problem Description
In a factory, there are N workers to finish two types of tasks (A and B). Each type has N tasks. Each task of type A needs xi time to finish, and each task of type B needs yj time to finish, now, you, as the boss of the factory, need to make an assignment, which makes sure that every worker could get two tasks, one in type A and one in type B, and, what's more, every worker should have task to work with and every task has to be assigned. However, you need to pay extra money to workers who work over the standard working hours, according to the company's rule. The calculation method is described as follow: if someone’ working hour t is more than the standard working hour T, you should pay t-T to him. As a thrifty boss, you want know the minimum total of overtime pay.
 
Input
There are multiple test cases, in each test case there are 3 lines. First line there are two positive Integers, N (N<=1000) and T (T<=1000), indicating N workers, N task-A and N task-B, standard working hour T. Each of the next two lines has N positive Integers; the first line indicates the needed time for task A1, A2…An (Ai<=1000), and the second line is for B1, B2…Bn (Bi<=1000).
 
Output
For each test case output the minimum Overtime wages by an integer in one line.
 
Sample Input
2 5
4 2
3 5
 
Sample Output
4
 
Source
 题意:(转)有两个长度为N(N<=1000)的序列A和B,把两个序列中的共2N个数分为N组,使得每组中的两个数分别来自A和B,每组的分数等于max(0,组内两数之和-t),问所有组的分数之和的最小值。
分析:贪心策略,将A,B排序,A中最大的和B中最小的一组,即将A升序排序,将B降序排序,这样便是最优解。。。
 #include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn =1e3+;
int a[maxn],b[maxn];
bool cmp(const int a,const int b){
return a>b;
}
int main(){
int n,t;
while(~scanf("%d%d",&n,&t)){
for( int i=; i<n; i++ ){
cin>>a[i];
}
for( int i=; i<n; i++ ){
cin>>b[i];
}
sort(a,a+n);
sort(b,b+n,cmp);
int ans=;
for(int i=; i<n; i++ ){
if(a[i]+b[i]>t) ans+=a[i]+b[i]-t;
}
cout<<ans<<endl;
}
return ;
}

Assignments---(贪心)的更多相关文章

  1. hdu 3661 Assignments (贪心)

    Assignments Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total ...

  2. HDU 3661 Assignments (水题,贪心)

    题意:n个工人,有n件工作a,n件工作b,每个工人干一件a和一件b,a[i] ,b[i]代表工作时间,如果a[i]+b[j]>t,则老板要额外付钱a[i]+b[j]-t;现在要求老板付钱最少: ...

  3. Assignments

    Assignments Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Tota ...

  4. German Collegiate Programming Contest 2013-B:Booking(贪心)

        Booking Pierre is in great trouble today! He is responsible for managing the bookings for the AC ...

  5. sgu 195 New Year Bonus Grant【简单贪心】

    链接: http://acm.sgu.ru/problem.php?contest=0&problem=195 http://acm.hust.edu.cn/vjudge/contest/vi ...

  6. 【网络流+贪心】Homework

    题目描述 Taro is a student of Ibaraki College of Prominent Computing. In this semester, he takes two cou ...

  7. BZOJ 1692: [Usaco2007 Dec]队列变换 [后缀数组 贪心]

    1692: [Usaco2007 Dec]队列变换 Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 1383  Solved: 582[Submit][St ...

  8. HDOJ 1051. Wooden Sticks 贪心 结构体排序

    Wooden Sticks Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) To ...

  9. HDOJ 1009. Fat Mouse' Trade 贪心 结构体排序

    FatMouse' Trade Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) ...

随机推荐

  1. 2012服务器在IIS部署的SLL(https)网址谷歌浏览器无法访问的问题解决

    服务器环境:Windows Server 2012,IIS8. 当绑定了https,使用IE和Firefox浏览器能够正常访问,但是使用谷歌浏览器会出现net::ERR_CONNECTION_ABOR ...

  2. 【转】fiddler抓包HTTPS请求

    本文转自:http://blog.csdn.net/idlear/article/details/50999490 fiddler抓包HTTPS请求 跟着教程来,保证100%成功抓HTTPS包 教程开 ...

  3. grid - 网格项目层级

    网格项目可以具有层级和堆栈,必要时可能通过z-index属性来指定. 1.在这个例子中,item1和item2的开始行都是1,item1列的开始是1,item2列的开始是2,并且它们都跨越两列.两个网 ...

  4. ionic 状态栏显示异常 statusBar

    从主分支上新建一个分支开发另一个app, 生成之后手机上显示状态栏异常, 如下图, 只显示了电池的色块, 百思不得其解啊. 各种猜测无果, 对比config.xml, 发现statusBar插件版本不 ...

  5. 对Faster R-CNN的理解(2)

    2. 区域建议网络 区域建议网络(Regional Proposal Network, RPN),根据特征图上每一个点的向量,为这个点生成k个矩形建议框.每一个点输出的内容包括:reg层4个输出x.y ...

  6. Git回滚代码到某个commit

    回退命令: $ git reset --hard HEAD^ 回退到上个版本$ git reset --hard HEAD~3 回退到前3次提交之前,以此类推,回退到n次提交之前 $ git rese ...

  7. CentOS 7.4nginx配置SSL

    一.在/etc/nginx/conf.d目录下创建虚拟主机配置文件 server { listen 80; server_name www.xx.com xx.com; return 301 http ...

  8. Python中的retry

    1.通过语言特性实现: for i in range(0,100): while True: try: # do stuff except SomeSpecificException: continu ...

  9. matlab将矩阵数据归一化到[0,255]

    matlab将矩阵数据归一化到[0,255]     function OutImg = Normalize(InImg) ymax=255;ymin=0; xmax = max(max(InImg) ...

  10. 当我们用自定义的类对象作为key时,我们必须在程序中覆盖HashCode()和equals()。

    Key值既然可以作为对象,那么也可以用一个自定义的类.比如: m.put(new sutdent(“Liucy”,30),”boss”) 如果没有语句来判定Student类对象是否相同,则会全部打印出 ...