Hot Days Codeforces Round #132 (Div. 2) D(贪婪)
Description
The official capital and the cultural capital of Berland are connected by a single road running through n regions. Each region has a unique climate, so the i-th (1 ≤ i ≤ n) region
has a stable temperature of ti degrees in summer.
This summer a group of m schoolchildren wants to get from the official capital to the cultural capital to visit museums and sights. The trip organizers transport the children between the cities
in buses, but sometimes it is very hot. Specifically, if the bus is driving through the i-th region and has k schoolchildren, then the
temperature inside the bus is ti + k degrees.
Of course, nobody likes it when the bus is hot. So, when the bus drives through the i-th region, if it has more than Ti degrees
inside, each of the schoolchild in the bus demands compensation for the uncomfortable conditions. The compensation is as large as xi rubles and it is charged
in each region where the temperature in the bus exceeds the limit.
To save money, the organizers of the trip may arbitrarily add or remove extra buses in the beginning of the trip, and between regions (of course, they need at least one bus to pass any region). The organizers can also arbitrarily sort the children into buses,
however, each of buses in the i-th region will cost the organizers costi rubles. Please note
that sorting children into buses takes no money.
Your task is to find the minimum number of rubles, which the organizers will have to spend to transport all schoolchildren.
Input
The first input line contains two integers n and m(1 ≤ n ≤ 105; 1 ≤ m ≤ 106) —
the number of regions on the way and the number of schoolchildren in the group, correspondingly. Next n lines contain four integers each: the i-th
line contains ti, Ti, xi and costi (1 ≤ ti, Ti, xi, costi ≤ 106).
The numbers in the lines are separated by single spaces.
Output
Print the only integer — the minimum number of roubles the organizers will have to spend to transport all schoolchildren.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64dspecifier.
Sample Input
2 10
30 35 1 100
20 35 10 10
120
3 100
10 30 1000 1
5 10 1000 3
10 40 1000 100000
200065
Hint
In the first sample the organizers will use only one bus to travel through the first region. However, the temperature in the bus will equal30 + 10 = 40 degrees and each of 10 schoolchildren
will ask for compensation. Only one bus will transport the group through the second region too, but the temperature inside won't exceed the limit. Overall, the organizers will spend 100 + 10 + 10 = 120 rubles.
题意:n个城市。m个人。每个城市有一个温度t,车内的温度不能超过T,否者赔偿每个人x元。每一辆车的价格c,那个车内的温度为t加车内的人。对于每个站,你能够选择坐计量车,求花费最小?
思路:对于每个站,假设T<t 那么一辆车。假设t+m<=T,一辆车。假设t<T可是t+m>T;
那么就要考虑了
如今如果全部人在一辆车,价格为c+m*x;
假设有一部分人出去租车,发现用的钱少一点,那么那一辆车最好坐满,否者空出来的位置能够坐的人就在那个已经满的车内须要补偿,
假设一部分人出去发现划算。那么为什么在出去一部分呢?这样想来就仅仅有两种情况了
1‘ 全部人在一辆车上
2 每一个车都不超过温度T,可是都坐满了
好了,上代码了:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std; #define N 1000100 typedef __int64 ll; ll n,m,t,T,x,c; int main()
{
ll ans;
while(~scanf("%I64d%I64d",&n,&m))
{
ans=0;
while(n--)
{
scanf("%I64d%I64d%I64d%I64d",&t,&T,&x,&c);
if(T<=t)
{
ans+=x*m+c;
continue;
}
if(t+m<=T)
{
ans+=c;
continue;
}
ll temp=T-t;
temp=m%temp? m/temp+1:m/temp;
ans+=min(x*m+c,c*temp); }
printf("%I64d\n",ans);
}
return 0;
}
版权声明:本文博客原创文章,博客,未经同意,不得转载。
Hot Days Codeforces Round #132 (Div. 2) D(贪婪)的更多相关文章
- Codeforces Round #132 (Div. 2)
A. Bicycle Chain 统计\(\frac{b_j}{a_i}\)最大值以及个数. B. Olympic Medal \(\frac{m_{out}=\pi (r_1^2-r_2^2)hp_ ...
- Educational Codeforces Round 132 (Rated for Div. 2)
Educational Codeforces Round 132 (Rated for Div. 2) A. Three Doors 简述 题意: 有三扇门(1~3), 其中两扇门后面有对应标号门的钥 ...
- Codeforces Round #366 (Div. 2) ABC
Codeforces Round #366 (Div. 2) A I hate that I love that I hate it水题 #I hate that I love that I hate ...
- Codeforces Round #354 (Div. 2) ABCD
Codeforces Round #354 (Div. 2) Problems # Name A Nicholas and Permutation standard input/out ...
- Codeforces Round #368 (Div. 2)
直达–>Codeforces Round #368 (Div. 2) A Brain’s Photos 给你一个NxM的矩阵,一个字母代表一种颜色,如果有”C”,”M”,”Y”三种中任意一种就输 ...
- cf之路,1,Codeforces Round #345 (Div. 2)
cf之路,1,Codeforces Round #345 (Div. 2) ps:昨天第一次参加cf比赛,比赛之前为了熟悉下cf比赛题目的难度.所以做了round#345连试试水的深浅..... ...
- Codeforces Round #279 (Div. 2) ABCDE
Codeforces Round #279 (Div. 2) 做得我都变绿了! Problems # Name A Team Olympiad standard input/outpu ...
- Codeforces Round #262 (Div. 2) 1003
Codeforces Round #262 (Div. 2) 1003 C. Present time limit per test 2 seconds memory limit per test 2 ...
- Codeforces Round #262 (Div. 2) 1004
Codeforces Round #262 (Div. 2) 1004 D. Little Victor and Set time limit per test 1 second memory lim ...
随机推荐
- Centos系统各种日志存详解
Centos系统各种日志存储路径和详细介绍 Linux常见的日志文件详述如下 1./var/log/boot.log(自检过程) 2./var/log/cron (crontab守护进程crond所派 ...
- hdu 4706
注意一点 空的地方打空格而不是空字符,我因为这wa了一次... #include<cstdio> #include<cstring> #include<cstdlib&g ...
- 什么是防盗链设置中的空Referer
设置防盗链时候指明和不指明空Referer的差别及实现后的效果? 什么是Referer? 这里的 Referer 指的是HTTP头部的一个字段,也称为HTTP来源地址(HTTP Referer).用来 ...
- IOS之【地图MapKit】
iOS地图位置开发 iPhone SDK提供了三个类来管理位置信息:CLLocation CLLocationManager 和 CLLHeading(不常用).除了使用GPS来获取当前的位置信息 ...
- 与众不同 windows phone (10) - Push Notification(推送通知)之推送 Tile 通知, 推送自定义信息
原文:与众不同 windows phone (10) - Push Notification(推送通知)之推送 Tile 通知, 推送自定义信息 [索引页][源码下载] 与众不同 windows ph ...
- 5350.support
3G6200N3G6200NL3G300MAIR3GIIALL02393GALL0256NALL5002ALL5003ARGUS_ATP52B,ASL26555AWM002EVBAWAPN2403BC ...
- Swift - 运算符重载和运算符函数
让已有的运算符对自定义的类和结构进行运算或者重新定义已有运算符的运算规则,这种机制被称为运算符重载. 1,通过重载加号运算符,使自定义的两个坐标结构体对象实现相加: 1 2 3 4 5 6 7 8 9 ...
- Oracle 12C 简介
2013年6月26日,Oracle Database 12c版本正式发布,首先发布的版本号是12.1.0.1.0,率先提供下载的平台有Linux和Solaris: Oracle官方下载地址: http ...
- Android性能优化---布局优化
我们从事Android开发编写布局的时候大多数是使用XML来布局,这给我们带来了方便性,这样操作可以布局界面的代码和逻辑控制的Java代码分离出来,使程序的结构更加清晰.明了.特别的复杂的布局,但是这 ...
- ORA-00376:file x cannot be read at this time
之前出现过机房断电情况,重启数据库后发现出现ORA-00376的错误. 通过查询数据文件状态: SQL> select file_id,online_status from dba_data_f ...