解题报告

题意,有n个机器。m个任务。

每一个机器至多能完毕一个任务。对于每一个机器,有一个最大执行时间Ti和等级Li,对于每一个任务,也有一个执行时间Tj和等级Lj。仅仅有当Ti>=Tj且Li>=Lj的时候,机器i才干完毕任务j,并获得500*Tj+2*Lj金钱。

问最多能完毕几个任务,当出现多种情况时,输出获得金钱最多的情况。

对任务和机器进行从大到小排序,从最大时间且最大等级的任务開始,选取全部符合时间限制的机器中等级要求最低的,这样能保证以下任务可选择的机器最多,如果选择的机器中等级要求最大的,接下去有一任务同样时间。却可能找不到可用的机器。

如机器(200,5)(200,2)任务(100,2)(99,4)

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define N 100010
#define LL __int64
using namespace std;
struct node
{
int t,l;
} ma[N],ta[N];
int n,m;
int cmp(node a,node b)
{
if(a.t==b.t)
return a.l>b.l;
else return a.t>b.t;
}
int main()
{
int i,j,k;
while(~scanf("%d%d",&n,&m))
{
LL sum=0;
int cnt=0;
for(i=0; i<n; i++)
scanf("%d%d",&ma[i].t,&ma[i].l);
for(i=0; i<m; i++)
scanf("%d%d",&ta[i].t,&ta[i].l);
sort(ma,ma+n,cmp);
sort(ta,ta+m,cmp);
int cc[110];
memset(cc,0,sizeof(cc));
for(i=0,j=0; i<m; i++)
{
while(j<n&&ta[i].t<=ma[j].t)
{
cc[ma[j].l]++;
j++;
}
for(k=ta[i].l; k<=100; k++)
{
if(cc[k])
{
sum+=(500*ta[i].t+2*ta[i].l);
cnt++;
cc[k]--;
break;
}
}
}
printf("%d %I64d\n",cnt,sum);
}
return 0;
}

Task

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

Total Submission(s): 1297    Accepted Submission(s): 321

Problem Description
Today the company has m tasks to complete. The ith task need xi minutes to complete. Meanwhile, this task has a difficulty level yi. The machine whose level below this task’s level yi cannot complete this task. If the company completes this task, they will
get (500*xi+2*yi) dollars.

The company has n machines. Each machine has a maximum working time and a level. If the time for the task is more than the maximum working time of the machine, the machine can not complete this task. Each machine can only complete a task one day. Each task
can only be completed by one machine.

The company hopes to maximize the number of the tasks which they can complete today. If there are multiple solutions, they hopes to make the money maximum.
 
Input
The input contains several test cases. 

The first line contains two integers N and M. N is the number of the machines.M is the number of tasks(1 < =N <= 100000,1<=M<=100000).

The following N lines each contains two integers xi(0<xi<1440),yi(0=<yi<=100).xi is the maximum time the machine can work.yi is the level of the machine.

The following M lines each contains two integers xi(0<xi<1440),yi(0=<yi<=100).xi is the time we need to complete the task.yi is the level of the task.
 
Output
For each test case, output two integers, the maximum number of the tasks which the company can complete today and the money they will get.
 
Sample Input
1 2
100 3
100 2
100 1
 
Sample Output
1 50004
 
Author
FZU
 
Source
 

2014 Multi-University Training Contest 1/HDU4864_Task(贪心)的更多相关文章

  1. HDU4888 Redraw Beautiful Drawings(2014 Multi-University Training Contest 3)

    Redraw Beautiful Drawings Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Jav ...

  2. hdu 4946 2014 Multi-University Training Contest 8

    Area of Mushroom Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) ...

  3. 2014 Multi-University Training Contest 9#11

    2014 Multi-University Training Contest 9#11 Killing MonstersTime Limit: 2000/1000 MS (Java/Others)   ...

  4. 2014 Multi-University Training Contest 9#6

    2014 Multi-University Training Contest 9#6 Fast Matrix CalculationTime Limit: 2000/1000 MS (Java/Oth ...

  5. 2014 Multi-University Training Contest 1/HDU4861_Couple doubi(数论/法)

    解题报告 两人轮流取球,大的人赢,,, 贴官方题解,,,反正我看不懂.,,先留着理解 关于费马小定理 关于原根 找规律找到的,,,sad,,, 非常easy找到循环节为p-1,每个循环节中有一个非零的 ...

  6. 2017 Multi-University Training Contest - Team 1 1002&&HDU 6034 Balala Power!【字符串,贪心+排序】

    Balala Power! Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)T ...

  7. 2018 Multi-University Training Contest 1 Distinct Values 【贪心 + set】

    任意门:http://acm.hdu.edu.cn/showproblem.php?pid=6301 Distinct Values Time Limit: 4000/2000 MS (Java/Ot ...

  8. hdu 4937 2014 Multi-University Training Contest 7 1003

    Lucky Number Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) T ...

  9. hdu 4941 2014 Multi-University Training Contest 7 1007

    Magical Forest Time Limit: 24000/12000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Other ...

随机推荐

  1. 如何写科技论文How to write a technical paper

    This is the evolving set of recommendations I share with my graduate students for technical writing. ...

  2. Java里面类型转换总结

    1.String 转 int int i = Integer.valueOf(my_str).intValue(); int i = Integer.parseInt(str); 2.String 转 ...

  3. 类的const和非const成员函数的重载

    我们从一个例子说起,来看上一篇文章中的String类, 我们为它提供一个下标操作符([ ])以读写指定位置的字符(char). 只要了解过C++的操作符重载的语法,很快就可以写出下面这个[]操作符重载 ...

  4. 那些年困扰我们的Linux 的蠕虫、病毒和木马

    虽然针对Linux的恶意软件并不像针对Windows乃至OS X那样普遍,但是近些年来,Linux面临的安全威胁却变得越来越多.越来越严重.个中原因包括,手机爆炸性的普及意味着基于Linux的安卓成为 ...

  5. KVM&amp;Libvirt基本概念及开发杂谈

    导读 大家好,本次肖力分享的主题是KVM&Libvirt基本概念及开发杂谈,内容有些凌乱松散,主要基于自己早期整理的笔记内容和实践感悟,有些内容难免有失偏颇,望见谅.前面先介绍下需要了解的基本 ...

  6. RS报表设计采用Total汇总过滤出错

    错误信息: DMR 子查询计划失败,并产生意外错误.: java.lang.NullPointerException 如图 原因是在RS过滤器中添加了: total([门诊人次] for [明细科室] ...

  7. HDU 5025图论之BFS

    点击打开链接 题意:从K走到T,S为怪,走的时候就多花费一秒,走到T时收集m把不同的钥匙.可是规定收集n之前,必须1~n-1所有收集完成,怪最多有5个 思路:怪最多就有5个,然后钥匙是1~9把,我们每 ...

  8. Android studio DrawerLayout

    网上开源项目地址:https://github.com/ikimuhendis/LDrawer 效果图: watermark/2/text/aHR0cDovL2Jsb2cuY3Nkbi5uZXQvQW ...

  9. C#应用视频教程2.1 OPENGL虚拟仿真介绍

    OPENGL的虚拟仿真对于工控自动化的意义很大,虽然市面上有很多的第三方软件比如Solidworks,Mathlab,ProE等等软件可以做仿真,而且能够实现的功能包括了流体分析,力学分析,摩擦力分析 ...

  10. zoj3662Math Magic

    Math Magic Time Limit: 3 Seconds       Memory Limit: 32768 KB Yesterday, my teacher taught us about ...