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. 倍福TwinCAT(贝福Beckhoff)常见问题(FAQ)-如何在同一台PC上运行多个TwinCAT程序

    右击桌面右下角的TC2图标,切换到PLC Configuration,然后在Plc Settings中设置数量为4(TC2最多可以运行的数量是4个),然后点击Apply   可能需要输入登录用户名和密 ...

  2. 正则表达式学习(PCRE)

    正则表达式是一个从左到右匹配目标字符串的模式.大多数字符自身就代表一个匹配 它们自身的模式. 1.分隔符:当使用 PCRE 函数的时候,模式需要由分隔符闭合包裹.分隔符可以使任意非字母数字.非反斜线. ...

  3. 演示程序之打游戏 -- 慕司板IAP15

    上位机和协议制定我的大学舍友(他的微博:http://weibo.com/lesshst? topnav=1&wvr=5&topsug=1)毕业前百忙之中使用Python花了一个下午完 ...

  4. js实现全选,全不选,反选

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

  5. Thinkphp+AJAX动态验证用户输入是否合法

    遇到用户注冊等情况时.假设等用户输入全部信息,点击注冊button提交后.再验证输入是否正确,体验非常不好,并且非常浪费用户的时间,添加注冊成本,这里提供一个样例,演示了怎么使用ajax进行单步验证, ...

  6. blender, fbx导入blender进行编辑

    fbx文件导入blender后,直接点下面Object Mode弹不出下拉菜单,从而无法进入Edit Mode.解法是先点一下右边Scene层级列表中的Sphere节点,将其选中,然后再点下面的Obj ...

  7. DataProtectionConfigurationProvider加密web.config文件

    web.config 文件中经常会包含一些敏感信息,最常见的就是数据库连接字符串了,为了防止该信息泄漏,最好是将相关内容加密. Aspnet_regiis.exe命令已经提供了加密配置文件的方法,系统 ...

  8. [个人开发人员赚钱九]做一个日收入10元的APP!

    [导语]尽管讲了非常多个人开发人员的文章.但新手开发人员怎样赚自己的第一个10块钱.确是最难的事情.群里有人说都不知道干什么app赚钱.全然没有想法.而且常常问我有什么高速赚钱的方法.我仅仅能遗憾地 ...

  9. makefile之origin函数

    origin 函数的作用是告诉你变量是哪里来的,其出生状况如何,他并不改变变量. 函数语法: $(origin ) 为变量的名字,而不是引用,所以一般没有"$"字符在前. orig ...

  10. 日期在苹果手机上显示NaN的处理方法

    注意两点即可: 1.苹果只认识 yyyy/mmmm/dddd/  这类格式的日期 2.如果输出后还要进行处理日期对比,苹果默认会带中文字,如:年月日,需要转成上面1当中的日期格式在转时间戳进行比较 G ...