POJ:2393-Yogurtfactory 编程题

Yogurt factory

Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 13073 Accepted: 6577

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

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.

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解题心得：

题意就是一家工厂，每天生产一单位牛奶的单价是s，每天必须交货y单位的牛奶，如果储存牛奶，储存的牛奶的单价每天会上涨c。问如果按要求提供牛奶，最少花费是多少。 按照贪心，可以假设将每天的牛奶都储存起来了，但是暂时不算储存的钱，只有每次取最小花费牛奶的时候算钱。但是怎么维护所有牛奶储存起来的信息呢？其实不用维护所有的信息，只要维护最小价格的那个牛奶就行了，因为你每次都是只取最小的，而所有的储存牛奶都是每次涨c元最小的值只会在新的一天的时候变换，所以直接维护最小值，每次用最小值来进行计算。

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#include 
#include 
using namespace std;
typedef long long ll;
int main() {
    ll n,s,Min = 0x7f7f7f7f;
    ll ans = 0;
    scanf("%lld%lld",&n,&s);
    for(int i=0;i