A supermarket has a set Prod of products on sale. It earns a profit px for each product x∈Prod sold by a deadline dx that is measured as an integral number of time units starting from the moment the sale begins. Each product takes precisely one unit of time for being sold. A selling schedule is an ordered subset of products Sell ≤ Prod such that the selling of each product x∈Sell, according to the ordering of Sell, completes before the deadline dx or just when dx expires. The profit of the selling schedule is Profit(Sell)=Σ x∈Sellpx. An optimal selling schedule is a schedule with a maximum profit. 
For example, consider the products Prod={a,b,c,d} with (pa,da)=(50,2), (pb,db)=(10,1), (pc,dc)=(20,2), and (pd,dd)=(30,1). The possible selling schedules are listed in table 1. For instance, the schedule Sell={d,a} shows that the selling of product d starts at time 0 and ends at time 1, while the selling of product a starts at time 1 and ends at time 2. Each of these products is sold by its deadline. Sell is the optimal schedule and its profit is 80. 

Write a program that reads sets of products from an input text file and computes the profit of an optimal selling schedule for each set of products. 

Input

A set of products starts with an integer 0 <= n <= 10000, which is the number of products in the set, and continues with n pairs pi di of integers, 1 <= pi <= 10000 and 1 <= di <= 10000, that designate the profit and the selling deadline of the i-th product. White spaces can occur freely in input. Input data terminate with an end of file and are guaranteed correct.

Output

For each set of products, the program prints on the standard output the profit of an optimal selling schedule for the set. Each result is printed from the beginning of a separate line.

Sample Input

4  50 2  10 1   20 2   30 1

7  20 1   2 1   10 3  100 2   8 2
5 20 50 10

Sample Output

80
185

Hint

The sample input contains two product sets. The first set encodes the products from table 1. The second set is for 7 products. The profit of an optimal schedule for these products is 185.
 
 
 
题目大意是买卖N件东西,每件东西都有个截止时间,在截止时间之前买都可以,
而每个单位时间只能买一件。问最大获利。
这题是一个贪心水题,进行价格排序,如果当天已经被选则推至前一天 
 
 
 #include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#include<cctype>
using namespace std;
struct node
{
int x,y;
}qu[];
int vis[];
int cmp(node a,node b) {
return a.x>b.x;
}
int main() {
int n;
while(scanf("%d",&n)!=EOF){
memset(vis,,sizeof(vis));
for (int i= ;i<n ;i++)
scanf("%d%d",&qu[i].x,&qu[i].y);
sort(qu,qu+n,cmp);
int temp=,sum=;
for (int i= ;i<n ;i++){
if (vis[qu[i].y]==) {
vis[qu[i].y]=;
sum+=qu[i].x;
}else {
for (int j=qu[i].y- ;j>= ;j--){
if (vis[j]==) {
sum+=qu[i].x;
vis[j]=;
break;
}
}
}
}
printf("%d\n",sum);
}
return ;
}
 

Supermarket POJ - 1456的更多相关文章

  1. (并查集 贪心思想)Supermarket -- POJ --1456

    链接: http://poj.org/problem?id=1456 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=82830#probl ...

  2. Day5 - H - Supermarket POJ - 1456

    A supermarket has a set Prod of products on sale. It earns a profit px for each product x∈Prod sold ...

  3. Supermarket POJ - 1456 贪心+并查集

    #include<iostream> #include<algorithm> using namespace std; const int N=1e5; struct edge ...

  4. Supermarket POJ - 1456(贪心)

    题目大意:n个物品,每个物品有一定的保质期d和一定的利润p,一天只能出售一个物品,问最大利润是多少? 题解:这是一个贪心的题目,有两种做法. 1 首先排序,从大到小排,然后每个物品,按保质期从后往前找 ...

  5. POJ 1456 Supermarket 区间问题并查集||贪心

    F - Supermarket Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Sub ...

  6. POJ 1456——Supermarket——————【贪心+并查集优化】

    Supermarket Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit  ...

  7. POJ 1456 - Supermarket - [贪心+小顶堆]

    题目链接:http://poj.org/problem?id=1456 Time Limit: 2000MS Memory Limit: 65536K Description A supermarke ...

  8. poj 1456 Supermarket - 并查集 - 贪心

    题目传送门 传送点I 传送点II 题目大意 有$n$个商品可以销售.每个商品销售会获得一个利润,但也有一个时间限制.每个商品需要1天的时间销售,一天也只能销售一件商品.问最大获利. 考虑将出售每个物品 ...

  9. POJ 1456 Supermarket(贪心+并查集)

    题目链接:http://poj.org/problem?id=1456 题目大意:有n件商品,每件商品都有它的价值和截止售卖日期(超过这个日期就不能再卖了).卖一件商品消耗一个单位时间,售卖顺序是可以 ...

随机推荐

  1. BZOJ 1202: [HNOI2005]狡猾的商人 [带权并查集]

    题意: 给出m个区间和,询问是否有区间和和之前给出的矛盾 NOIp之前做过hdu3038..... 带权并查集维护到根的权值和,向左合并 #include <iostream> #incl ...

  2. ThinkPHP删除栏目(多)

    前段时间发表了一个删除栏目的随笔,当时实现的功能是删除一条信息,这次来实现一下批量删除栏目. 我们需要达到的是这样一个效果: 选中批量删除按钮后可以选中所有该页面的栏目,这个是前端页面的实现,在这里就 ...

  3. console那些你不曾知道的玩法

    一.console最常见的四种方法: FireFox(58) Chrome(51) 二.打印对象: 平时想输出对象属性时,可以直接打印对象,对Object使用toString方法会得到 [Object ...

  4. HTTPS的原理解析

    http://www.cnblogs.com/alisecurity/p/5939336.html 外加文档

  5. [bzoj]2962序列操作

    [bzoj]2962序列操作 标签: 线段树 题目链接 题意 给你一串序列,要你维护三个操作: 1.区间加法 2.区间取相反数 3.区间内任意选k个数相乘的积 题解 第三个操作看起来一脸懵逼啊. 其实 ...

  6. 为何要部署IPV6

    ·IPv4的局限性:   1.地址空间的局限性:IP地址空间的危机由来已久,并正是升级到IPv6的主要动力.   2.安全性:IPv4在网络层没有安全性可言,安全性一直被认为是由网络层以上的层负责. ...

  7. JavaScript 知识点

    JS基础 页面由三部分组成: html:超文本标记语言,负责页面结构 css:层叠样式表,负责页面样式 js:轻量级的脚本语言,负责页面的动效和数据交互 小总结:结构,样式和行为,三者相分离 在htm ...

  8. Java.lang.Comparable接口和Java.util.Comparator接口的区别

    Java的Comparator和Comparable当需要排序的集合或数组不是单纯的数字型时,通常可以使用Comparator或Comparable,以简单的方式实现对象排序或自定义排序. 1.Com ...

  9. POJ1639 - Picnic Planning

    原题链接 Description 给出一张个点的无向边权图并钦定点,求使得点的度不超过的最小生成树. Solution 首先无视掉与相连的所有边,原图会变成若干互不连通的个块.对每个块分别求MST,再 ...

  10. 关于本地化(localization)

    关于本地化(localization) 我们都知道,如果不需要做国际化版本的App.我们只需要在info.plist 里修改CFBundleDisplayName就可以了,其实做国际化也就是在不同的国 ...