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. OpenCV学习(9) 分水岭算法(3)

    本教程我学习一下opencv中分水岭算法的具体实现方式. 原始图像和Mark图像,它们的大小都是32*32,分水岭算法的结果是得到两个连通域的轮廓图. 原始图像:(原始图像必须是3通道图像) Mark ...

  2. SRM 212 Div II Level Two: WinningRecord,Brute Force

    题目来源:http://community.topcoder.com/stat?c=problem_statement&pm=3003&rd=5858 比较简单. 代码如下: #inc ...

  3. Android内存调试命令

    adb shell dumpsys meminfo 包名 比如: adb shell dumpsys meminfo cn.com.test

  4. FreeBSD与Linux十个本质上的区别

    Linux的标志是一只十分可爱的小企鹅,而FreeBSD的标志是一个拿着叉子的小恶魔.你是否经常会听到人们把 Linux及 BSD 系统混为一谈?是的,我有时会经常听到一些新手,甚至于媒体都这么说.当 ...

  5. 切线空间(Tangent Space)法线映射(Normal Mapping)【转】

    // 请注明出处:http://blog.csdn.net/BonChoix,谢谢~) 切线空间(Tangent Space) 切换空间,同局部空间.世界空间等一样,是3D图形学中众多的坐标系之一.切 ...

  6. SVN-如何删除 SVN 文件夹下面的小图标

    第一步:右键新建文本文件. 第二步:打开 并复制一下文件 到 新建文本文件. @echo on    color 2f    mode con: cols=80 lines=25    @REM    ...

  7. 删除android ScrollView边界阴影方法

    XML文件中添加以下方法:   android:fadingEdge=”none”   或者,代码中设置为false即可   ScrollView.setHorizontalFadingEdgeEna ...

  8. UNIX网络编程读书笔记:readv和writev函数

    这两个函数类似于read和write,不过readv和writev允许单个系统调用读入到或写出自一个或多个缓冲区.这些操作分别称为分散读(scatter read)和集中写(gather write) ...

  9. CSV 读写

    using System; using System.IO; using System.Runtime.InteropServices; using UnityEngine; public class ...

  10. PotPlayer 如何设置多屏幕全屏播放

      如何在播放器中,设置扩展播放模式? 全屏设置/主全屏显示设备:Display2   如何使视频播放时,没有黑边并且全屏充满? 高宽比/处理方式:保持全屏宽高比   如果取消掉视频上方的文字提示信息 ...