C - Bolero

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Description

Winter in Yekaterinburg is the longest time of the year. And everyone spends long winter evenings in his own way. Denis is a big fan of classical music, at least twice a week he attends concerts at the Philharmonic.
This year, Denis wants to buy tickets for the whole season at once and even already has chosen a list of concerts, which he wants to attend. Now, he thinks, how to buy tickets. There is an opportunity in the Philharmonic to choose any k or more concerts and to buy a single subscription for all of them. The cost of this subscription will be equal to the total cost of all concerts in it but with a discount ppercent. There are many types of such subscriptions, with different k and p. In addition, for some concerts Denis, as a student, has a discount. This discount is valid only if he buys the ticket separately, outside the subscription.
Denis, like any other poor student, wants to attend all the concerts, spending as little money as possible. However, he knows that there are always more people wanting to attend a concert than the number of seats in the Philharmonic hall. So he will not buy more than one ticket for one concert (separately or in a subscription), even if it allows him to save money.

Input

The first line contains integers n and m that are the number of concerts Denis wants to attend, and a number of different subscription types at the Philharmonic (2 ≤ n ≤ 10 5; 1 ≤ m ≤ 10 5). The i-th of the following n lines contains integers si and di that are the price of a ticket and the discount for students in percent at the i-th concert (100 ≤ si ≤ 50 000; 0 ≤ di ≤ 100). The  j-th of the following m lines contains integers kj and pj that are the minimum number of concerts that can be combined in the subscription of the j-th type, and a discount for it in percent (2 ≤ kj ≤ n; 1 ≤ pj ≤ 100).

Output

Output the minimum amount of money Denis has to spend to attend all the concerts. The answer should be given with an absolute or relative error not exceeding 10 −6.

Sample Input

input output
6 2
500 0
700 0
300 0
400 0
500 50
800 0
5 10
6 15
2680.00

Notes

In the example, Denis needs to buy a subscription of the first type (with 10% discount) for all concerts, except the fifth one, and to buy a ticket for the fifth concert with a student discount of 50%.
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <cstring>
#include <string>
#include <algorithm>
using namespace std;
typedef long long ll; #define MM(a,b) memset(a,b,sizeof(a));
#define inf 0x7f7f7f7f
#define FOR(i,n) for(int i=1;i<=n;i++)
#define CT continue;
#define PF printf
#define SC scanf
const int maxn=100000+10; struct node{
ll val;int id;
bool operator<(const node &a) const{
return this->val>a.val;
}
};
vector<int> G[105];
priority_queue<node> q;
int num[105],g[105];
int main()
{
int n,m;
while(~SC("%d%d",&n,&m)){
ll ori=0,ans=0;
for(int i=0;i<=100;i++) G[i].clear();
FOR(i,n) {
int cost,p;
SC("%d%d",&cost,&p);
ori+=cost*(100-p);
G[p].push_back(cost);
}
for(int i=0;i<=100;i++) sort(G[i].begin(),G[i].end());
MM(num,inf);
FOR(i,m){
int peo,dis;
SC("%d%d",&peo,&dis);
num[dis]=min(num[dis],peo);
}
// PF("111111\n");
ans=ori;
FOR(k,100){
if(num[k]>n) CT;
MM(g,0);
// PF("2222222\n");
ll tmp=ori,cnt=0;
for(int i=0;i<=k;i++)
for(int j=0;j<G[i].size();j++){
tmp+=G[i][j]*(i-k);cnt++;
}
// PF("3333333\n");
while(q.size()) q.pop();
for(int i=k+1;i<=100;i++) {
if(G[i].size()) q.push((node){G[i][0]*(i-k),i});
g[i]=1;
}
// PF("444444\n");
while(cnt<num[k]){
node u=q.top();q.pop();
tmp+=u.val;
cnt++;
if(g[u.id]<G[u.id].size()){
int r=g[u.id];
q.push((node){G[u.id][r]*(u.id-k),u.id});
g[u.id]++;
}
}
ans=min(ans,tmp);
}
PF("%.2f\n",(double)ans/100.00);
}
return 0;
}

  错点:如果直接枚举优惠券的优惠值大小k(0-100),然后,计算出每种票在此优惠券下的可优惠

价格,再sort排下序是100*1e5*log1e5是会超时的,所以需要优化一下,考虑买单个票的时候,打的折

<=k的必须要算(可优惠),对于>k的,对每种k,每次都只能选其中的一种(贪心),所以,可用优先队列将复杂度降到1e5log100,

注意==k的也必须要贪啊,,因为要让大与k的数量尽可能小

URAL 2092 Bolero 贪心的更多相关文章

  1. D - Binary Lexicographic Sequence URAL - 1081 (贪心)

    题目链接:https://cn.vjudge.net/contest/275079#problem/D 具体思路:首先,我们可以观察到1-n位数的种数连起来是一个很有规律的数列,然后我们开始倒序建立. ...

  2. 贪心 URAL 1303 Minimal Coverage

    题目传送门 /* 题意:最少需要多少条线段能覆盖[0, m]的长度 贪心:首先忽略被其他线段完全覆盖的线段,因为选取更长的更优 接着就是从p=0开始,以p点为标志,选取 (node[i].l < ...

  3. ural 1303 Minimal Coverage【贪心】

    链接: http://acm.timus.ru/problem.aspx?space=1&num=1303 http://acm.hust.edu.cn/vjudge/contest/view ...

  4. URAL 2019 Pair: normal and paranormal (贪心) -GDUT联合第七场

    比赛题目链接 题意:有n个人每人拿着一把枪想要杀死n个怪兽,大写字母代表人,小写字母代表怪兽.A只能杀死a,B只能杀死b,如题目中的图所示,枪的弹道不能交叉.人和怪兽的编号分别是1到n,问是否存在能全 ...

  5. ural 1303 Minimal Coverage(贪心)

    链接: http://acm.timus.ru/problem.aspx?space=1&num=1303 按照贪心的思想,每次找到覆盖要求区间左端点时,右端点最大的线段,然后把要求覆盖的区间 ...

  6. URAL 1203 Scientific Conference dp?贪心

    题目:click here 分明就是贪心怎么会在dp的专题 #include <bits/stdc++.h> using namespace std; typedef unsigned l ...

  7. URAL 1203 Scientific Conference(贪心 || DP)

    Scientific Conference 之前一直在刷计算几何,邀请赛连计算几何的毛都买见着,暑假这一段时间就做多校.补多校的题目.刷一下一直薄弱的DP.多校假设有计算几何一定要干掉-.- 题意:给 ...

  8. URAL 1995 Illegal spices 贪心构造

    Illegal spices 题目连接: http://acm.timus.ru/problem.aspx?space=1&num=1995 Description Jabba: Han, m ...

  9. URAL 2021 Scarily interesting! (贪心+题意)

    题意:给定两个队伍的每个人的得分,让你安排怎么比赛才能使得观众知道冠军的时间最长. 析:贪心,很简单,就是先开始总分高的先出最差劲的,总分低的先出最厉害的,这个题当时实在是读的不明白啊,WA了好多次. ...

随机推荐

  1. 查询集 QuerySet

    1 概念 Django的ORM中存在查询集的概念. 查询集,也称查询结果集.QuerySet,表示从数据库中获取的对象集合. 当调用如下过滤器方法时,Django会返回查询集(而不是简单的列表): a ...

  2. JSON和AJAX基础

    前一段时间做老师留的企业图谱作业,和查询功能都需要用到AJAX .然后做爬虫的时候发现好多网站都用到的是页面的局部刷新,也就是发送的AJAX请求.就去学了一下.简单总结 什么是 JSON ? JSON ...

  3. 微信小程序上传图片及本地测试

    前端(.wxml) <view id="view1"> <view id="btns"> <image id="ima1 ...

  4. mvc布局(一)

    negut添加Optimization @System.Web.Optimization.Styles.Render( "~/Content/styles/css/font-awesome. ...

  5. C#int类型 转Byte[]

    方法1:使用左移和右移 int转化为byte[]:   public  byte[] intToBytes(int value)         {             byte[] src = ...

  6. [Android] Installation failed due to: ''pm install-create -r -t -S 4590498' returns error 'UNSUPPORTED''

    小米特有问题 关闭开发者选项的启用MIUI优化 不得不说, 这是真的坑...

  7. 从零开始搭建react应用

    用create-react-app搭建react应用,了解npm run start的工作过程. 第一步:安装脚手架 create-react-app 1. 在node里 npm install cr ...

  8. 帝国cms 权限操作

    <? if ($classid==5 || $classid==6 || $classid==7 || $classid==8 || $classid==9 || $classid==10 || ...

  9. vue 节流

    <!doctype html> <html lang="en"> <head> <meta charset="UTF-8&quo ...

  10. Python单元测试框架unittest重要属性 与 用例编写思路

    前言 本文为转载,原文地址作者列举python unittest这个测试框架的主要属性和 测试用例思路 unittest单元测试框架不仅可以适用于单元测试,还可以适用WEB自动化测试用例的开发与执行, ...