POJ_2393_Yogurt factory

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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.


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大致意思是给出生产成本和储存成本，要你找出完成上级目标的最佳生产方式。

dp  这周的最优单价是上周最优价加上储存成本和这周生产单价取min      dp[i] = min ( dp[i-1]+s , c[i] )        //一定要好好学数学T^T

#include<math.h>#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int main(){    long long n,s,ans=0;    long long c[10010],y[10010];    scanf ("%I64d%I64d",&n,&s);    for (int i=0;i<n;i++)        scanf ("%I64d%I64d",&c[i],&y[i]);    for (int i=1;i<n;i++)        c[i]=min(c[i-1]+s,c[i]);    for (int i=0;i<n;i++)        ans+=c[i]*y[i];    printf ("%I64d\n",ans);    return 0;}