Installing Apps Kattis - installingapps (贪心 + 背包)
Installing Apps Kattis - installingapps
Sandra recently bought her first smart phone. One of her friends suggested a long list of applications (more commonly known as “apps”) that she should install on the phone. Sandra immediately started installing the apps from the list, but after installing a few, the phone did not have enough disk space to install any more apps. Sometimes, the app installation failed because there was not even enough space to download the installation package. Other apps could be downloaded just fine, but had insufficient space to store the installed app.
Each app that Sandra installs has a download size dd and a storage size ss. To download the app, Sandra’s phone must have at least dd megabytes of free disk space. After the app has been installed, it then uses ss megabytes of disk space on the phone. The download size may be smaller than the storage size (e.g., if the app data is heavily compressed) or larger than the storage size (e.g., if the download contains material that might not get used such as translations to different languages). The installer is very efficient and can transform the downloaded package to an installed app without using any extra disk space. Thus, to install an app, the phone must have at least max(d,s)max(d,s) megabytes of free disk space.
Sandra quickly realised that she may have run out of space just because she installed apps in the wrong order. Thus, she decided to give the installation another try. She uninstalled all apps, and will now choose an installation order that lets her install the largest number of apps from the list. Sandra may not install any app more than once.
Help her determine what apps on the list she should install, and in what order.
Input
The input consists of:
One line with two integers nn, cc (1≤n≤500,1≤c≤100001≤n≤500,1≤c≤10000), the number of available apps and the available disk space of the phone in megabytes.
nn lines, each with two integers d,sd,s (1≤d,s≤100001≤d,s≤10000), the download size and storage size of an app, in megabytes.
Output
Output one line with the maximum number of apps that can be installed. Then output one line listing the numbers of those apps, in the order that Sandra should install them. In the case that no apps can be installed, this line can be omitted.
The apps are numbered from 11 to nn, in the order they are given in the input. If there are multiple optimal solutions, output any one of them.
Sample Input 1 | Sample Output 1 |
---|---|
2 100 |
2 |
Sample Input 2 | Sample Output 2 |
---|---|
2 100 |
0 |
题意:一个内存为V 的手机可以安装一些app,每一个app有两个不同的数值,表示已开始下载的大小和安装完以后的大小(必须下载的和安装后的大小都比手机的剩余容量要小才可以放进去),问最多可以放进去几个app,和放进去背包的一些顺序
题解:这是一个背包问题,但是普通的背包拿去是没有顺序关系的,这道题拿去的顺序关系是有关系的,前面的一个的拿去是有可能会影响后面一个app的,所以要先贪心一下。具体的贪心操作是按照d-s的大小来排序,就是下载的和安装后的相减的值的从大到小排序。为什么要这样来贪心我的理解是容量的改变多的话如果放在后面,他由于改变的多了会更加容易影响。同时这道题在uva上好像不能ac,可能是uva的数据源出问题了
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<sstream>
#include<cmath>
#include<stack>
#include<cstdlib>
#include <vector>
#include<queue>
using namespace std; typedef long long ll;
const int MAXN=1e4+5;
const double EPS=1e-8;
const int INF=0x3f3f3f3f;
const int MOD = 1e9+7;
struct Node{
int d,s,id;
}a[MAXN];
int f[505][MAXN],n,V;
bool cmp(const Node a, const Node b){
return (a.d - a.s) > (b.d - b.s);
//return a.s < b.s;
}
int main(){ ios::sync_with_stdio(false);
while(cin >> n >> V)
{
for(int i=1;i<=n;i++)
{
cin >> a[i].d >> a[i].s;
a[i].id = i;
}
sort(a+1,a+n+1,cmp);
memset(f,0,sizeof(f));
for(int i=1;i<=n;i++)
{
for(int j=V;j>=0;j--)
{
f[i][j] = f[i-1][j];
if(V-(j-a[i].s) < a[i].d)
continue;
if(j < a[i].s)
continue;
f[i][j] = max(f[i-1][j], f[i-1][j-a[i].s]+1);
}
}
int Max = 0, v = 0;
for(int i=0;i<=V;i++)
{
if(f[n][i] > Max)
{
Max = f[n][i];
v = i;
}
}
cout << f[n][v] << "\n";
if(f[n][v] == 0) continue;
int q[MAXN],cnt = 0;
for(int i=n;i>=1&&v>0;i--)
{
if(f[i][v] == f[i-1][v-a[i].s]+1)
{
q[++cnt] = a[i].id;
v -= a[i].s;
}
}
for(int i=cnt;i>=1;i--)
{
cout << q[i] << " \n"[i==1] ;
}
}
return 0;
}
Installing Apps Kattis - installingapps (贪心 + 背包)的更多相关文章
- 【bzoj4922】[Lydsy六月月赛]Karp-de-Chant Number 贪心+背包dp
题目描述 给出 $n$ 个括号序列,从中选出任意个并将它们按照任意顺序连接起来,求以这种方式得到匹配括号序列的最大长度. 输入 第一行包含一个正整数n(1<=n<=300),表示括号序列的 ...
- HDU3466Proud Merchants(贪心&背包)
http://acm.hdu.edu.cn/showproblem.php?pid=3466 题目大意是说n个物品每个物品的花费是p,但是如果你现在的钱少于q就买不了这个物品,每个物品的价值是v,求有 ...
- HDU 5303 Delicious Apples(贪心 + 背包 2015多校啊)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5303 Problem Description There are n apple trees plan ...
- bzoj4922 [Lydsy1706月赛]Karp-de-Chant Number 贪心+背包
题目传送门 https://lydsy.com/JudgeOnline/problem.php?id=4922 题解 记录每一个串的没有匹配的右括号 \()\) 的数量为 \(a_i\),为匹配的左括 ...
- AT4120-[ARC096D]Sweet Alchemy【贪心,背包】
正题 题目链接:https://www.luogu.com.cn/problem/AT4120 题目大意 给出\(n\)个物品和一个容量\(m\),第\(i\)个物品体积为\(c_i\).除了第一个物 ...
- HDU 4003 [树][贪心][背包]
/* 大连热身A题 不要低头,不要放弃,不要气馁,不要慌张 题意: 给一棵树,每条边上有权值.给一个起点,放置n个机器人,要求使得任意一个节点至少被一个机器人经过. 每个机器人经过某条边时的代价为这条 ...
- POJ2392Space Elevator(贪心+背包)
Space Elevator Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 9970 Accepted: 4738 De ...
- 【贪心+背包】【HDU2546】【饭卡】
饭卡 Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submiss ...
- sdut2408 pick apples (贪心+背包)山东省第三届ACM省赛
版权声明:本文为博主原创文章.未经博主同意不得转载. https://blog.csdn.net/svitter/article/details/24642587 本文出自:http://blog.c ...
随机推荐
- CentOS7.4搭建GitLab
1.查看服务器环境 uname -a 2.下载安装包 [1]找到相应的最新版本的下载路径 网址:https://mirrors.tuna.tsinghua.edu.cn/gitlab-ce/yum/e ...
- Asp.NET MVC+WebAPI跨域调用
使用jQuery调用WebApi有时会遇到跨域的问题,今天介绍一种可以简单解决跨域问题的方法. 当我们跨域请求WebAPI的时候会提示以下信息: XMLHttpRequest cannot load ...
- Visual Studio 使用
目录结构 solution_dir Debug: 存放Debug版本信息的.exe Release: Release的.exe .sln: visual studio 项目文件 project_dir ...
- HDU 5501——The Highest Mark——————【贪心+dp】
The Highest Mark Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Other ...
- 一、基础知识 React API 一览
1.10 Hooks 参考文章:https://juejin.im/post/5be3ea136fb9a049f9121014 demo: /** * 必须要react和react-dom 16.7以 ...
- MySQL-5.6.30 (OpenLogic CentOS7.2)
平台: CentOS 类型: 虚拟机镜像 软件包: centos7.2 mysql5.6.30 basic software database linux open source 服务优惠价: 按服务 ...
- Java JDBC链接Oracle数据库
package com.test.test; import java.io.FileInputStream;import java.io.FileNotFoundException;import ja ...
- C++ vector类详解
转自http://blog.csdn.net/whz_zb/article/details/6827999 vector简介 vector是STL中最常见的容器,它是一种顺序容器,支持随机访问.vec ...
- python request下载文件时,显示进度以及网速
import requests import time def downloadFile(name, url): headers = {'Proxy-Connection':'keep-alive'} ...
- 移动端rem匹配
Rem是相对于根元素font-size大小的单位 记inphone5屏幕宽度是 320px font-size16px 1rem=16px <html> <head> ...