#POJ2393#Yogurt factory（贪心）

Yogurt factory

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 10860 Accepted: 5519

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

4 588 20089 40097 30091 500

Sample Output

126900

Hint

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. 

题意：

酸奶可以无限制储存，储存每天每单位酸奶储存单价一定，

给出每天生产酸奶的单价，及每天的需求量，求最小代价

这题的关键在于无限制地储存，那么在第 i 天，要生产的 x 单位酸奶的单价，

可以看做之前所有生产的单价加上储存的单价然后比最小，而这个是可以一直跟着天数走的

（单价小的加上再多一天的储存价还是小，以为其它的也都要加储存价）

Code:

StatusAcceptedTime32msMemory668kBLength616LangG++Submitted2017-07-13 11:38:08Shared

#include<iostream>#include<cstdio>#include<cstdlib>#include<algorithm>using namespace std;typedef long long LL;const int INF = 0x3f3f3f3f;int N, S;void getint(int & num){    char c;    int flg = 1;    num = 0;    while((c = getchar()) < '0' || c > '9')    if(c == '-')    flg = -1;    while(c >= '0' && c <= '9')    {    num = num * 10 + c - 48;    c = getchar();}    num *= flg;}int main(){getint(N), getint(S);int c, y, lst = INF;LL Ans = 0LL;for(int i = 1; i <= N; ++ i){getint(c), getint(y);c = min(c, lst + S);lst = c;Ans += c * y;}printf("%I64d\n", Ans);return 0;}