http://www.lydsy.com/JudgeOnline/problem.php?id=1680

看不懂英文。。

题意是有n天,第i天生产的费用是c[i],要生产y[i]个产品,可以用当天的也可以用以前的(多生产的)。每单位产品保存一天的费用是s。求最小费用

显然贪心,每次查找之前有没有哪一天保存到现在的价值最小,然后比较更新。。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
#define rep(i, n) for(int i=0; i<(n); ++i)
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
#define for2(i,a,n) for(int i=(a);i<(n);++i)
#define for3(i,a,n) for(int i=(a);i>=(n);--i)
#define for4(i,a,n) for(int i=(a);i>(n);--i)
#define CC(i,a) memset(i,a,sizeof(i))
#define read(a) a=getint()
#define print(a) printf("%lld", a)
#define dbg(x) cout << #x << " = " << x << endl
#define printarr(a, n, m) rep(aaa, n) { rep(bbb, m) cout << a[aaa][bbb]; cout << endl; }
inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<'0'||c>'9'; c=getchar()) if(c=='-') k=-1; for(; c>='0'&&c<='9'; c=getchar()) r=r*10+c-'0'; return k*r; }
inline const int max(const int &a, const int &b) { return a>b?a:b; }
inline const int min(const int &a, const int &b) { return a<b?a:b; }
long long n, ans, s, sum=~0u>>2; int main() {
read(n); read(s);
for1(i, 1, n) {
long long a=getint(), b=getint();
sum+=s;
if(sum>a) sum=a;
ans+=sum*b;
}
print(ans);
return 0;
}

Description

The cows have purchased a yogurt factory that makes world-famous Yucky Yogurt. Over the next N (1 <= N <= 10,000) weeks, the price of milk and labor will fluctuate weekly such that it will cost the company C_i (1 <= C_i <= 5,000) cents to produce one unit of yogurt in week i. Yucky's factory, being well-designed, can produce arbitrarily many units of yogurt each week. Yucky Yogurt owns a warehouse that can store unused yogurt at a constant fee of S (1 <= S <= 100) cents per unit of yogurt per week. Fortuitously, yogurt does not spoil. Yucky Yogurt's warehouse is enormous, so it can hold arbitrarily many units of yogurt. Yucky wants to find a way to make weekly deliveries of Y_i (0 <= Y_i <= 10,000) units of yogurt to its clientele (Y_i is the delivery quantity in week i). Help Yucky minimize its costs over the entire N-week period. Yogurt produced in week i, as well as any yogurt already in storage, can be used to meet Yucky's demand for that week.

Input

* Line 1: Two space-separated integers, N and S. * Lines 2..N+1: Line i+1 contains two space-separated integers: C_i and Y_i.

Output

* Line 1: Line 1 contains a single integer: the minimum total cost to satisfy the yogurt schedule. Note that the total might be too large for a 32-bit integer.

Sample Input

4 5
88 200
89 400
97 300
91 500

Sample Output

126900

OUTPUT DETAILS:

In week 1, produce 200 units of yogurt and deliver all of it. In
week 2, produce 700 units: deliver 400 units while storing 300
units. In week 3, deliver the 300 units that were stored. In week
4, produce and deliver 500 units.

HINT

Source

【BZOJ】1680: [Usaco2005 Mar]Yogurt factory(贪心)的更多相关文章

  1. BZOJ 1680 [Usaco2005 Mar]Yogurt factory:贪心【只用考虑上一个】

    题目链接:http://www.lydsy.com/JudgeOnline/problem.php?id=1680 题意: 在接下来的n周内,第i周生产一吨酸奶的成本为c[i],订单为y[i]吨酸奶. ...

  2. bzoj 1680: [Usaco2005 Mar]Yogurt factory【贪心】

    贪心,一边读入一边更新mn,用mn更新答案,mn每次加s #include<iostream> #include<cstdio> using namespace std; in ...

  3. BZOJ 1740: [Usaco2005 mar]Yogurt factory 奶酪工厂 贪心 + 问题转化

    Description The cows have purchased a yogurt factory that makes world-famous Yucky Yogurt. Over the ...

  4. BZOJ1680: [Usaco2005 Mar]Yogurt factory

    1680: [Usaco2005 Mar]Yogurt factory Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 106  Solved: 74[Su ...

  5. 1740: [Usaco2005 mar]Yogurt factory 奶酪工厂

    1740: [Usaco2005 mar]Yogurt factory 奶酪工厂 Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 119  Solved:  ...

  6. bzoj1680[Usaco2005 Mar]Yogurt factory*&&bzoj1740[Usaco2005 mar]Yogurt factory 奶酪工厂*

    bzoj1680[Usaco2005 Mar]Yogurt factory bzoj1740[Usaco2005 mar]Yogurt factory 奶酪工厂 题意: n个月,每月有一个酸奶需求量( ...

  7. BZOJ1740: [Usaco2005 mar]Yogurt factory 奶酪工厂

    n<=10000天每天Ci块生产一东西,S块保存一天,每天要交Yi件东西,求最少花多少钱. 这个我都不知道归哪类了.. #include<stdio.h> #include<s ...

  8. [Usaco2005 mar]Yogurt factory 奶酪工厂

    接下来的N(1≤N10000)星期中,奶酪工厂在第i个星期要花C_i分来生产一个单位的奶酪.约克奶酪工厂拥有一个无限大的仓库,每个星期生产的多余的奶酪都会放在这里.而且每个星期存放一个单位的奶酪要花费 ...

  9. POJ 2393 Yogurt factory 贪心

    Description The cows have purchased a yogurt factory that makes world-famous Yucky Yogurt. Over the ...

随机推荐

  1. Remove all your local git branches but keep master

    Sometimes after a sprint, all the remaining branches are just taking up space. Here's a small snippe ...

  2. react-redux 中 connect 的常用写法

    1.方式一 export default connect(({ monitorRedux })=>{ return { data: monitorRedux.data } })(Monitor) ...

  3. Openerp开发进销存系统总结

    转自 :http://blog.sina.com.cn/s/blog_7cb52fa80101ngt8.html 差不多用了2个星期的闲余事件,对于openerp v7.0进行了学习和应用开发.细节总 ...

  4. HBase - 计数器 - 计数器的介绍以及使用 | 那伊抹微笑

    博文作者:那伊抹微笑 csdn 博客地址:http://blog.csdn.net/u012185296 itdog8 地址链接 : http://www.itdog8.com/thread-215- ...

  5. Spring Boot(二)Application events and listeners

    一.自定义监听器: 1.创建: META-INF/spring.factories 2.添加: org.springframework.context.ApplicationListener=com. ...

  6. linux sudo命令详解

    --sudo命令重启网卡 sudo service network restart http://bestchenwu.iteye.com/blog/1450292

  7. Linux修改时间的方法

    http://www.blogjava.net/itvincent/archive/2007/08/03/134242.html修改linux的时间可以使用date指令 在命令行输入: date 显示 ...

  8. 取SQL分组中某几行数据

    常用的方法有:子查询.ROW_NUMBER.APPLY,总体感觉还是ROW_NUMBER比较直观 if OBJECT_ID('testGroup') is not null drop table te ...

  9. atitit..国富论 在现代it企业项目管理中的作用attialx 总结---国富论读后感 attialx

    atitit..国富论 在现代it企业项目管理中的作用attialx 总结---国富论读后感 attialx 1. 国民财富的性质和原因的研究(简称:<国富论>) 1 2. 蕴含的重要管理 ...

  10. Docker使用Dockerfile创建支持ssh服务自启动的容器镜像

    原文链接:Docker使用Dockerfile创建支持ssh服务自启动的容器镜像 1. 首先创建一个Dockerfile文件.文件内容例如以下 # 选择一个已有的os镜像作为基础 FROM cento ...