POJ2393 Yogurt factory (贪心)

Yogurt factory
Time Limit: 1000MS
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.
  Source
  USACO 2005 March Gold

#include<cstdio>#include<algorithm>#include<iostream>#include<cmath>#include<iomanip>#include<cstring>#include<vector>#include<iterator>#define N 10001using namespace std;int min(int a, int b){    return a<b?a:b;}int main(){    int n,s,c[N],y[N],i;    long long sum;    while(scanf("%d %d", &n,&s)!=EOF)    {        sum=0;        for(i=0; i<n; i++){            scanf("%d %d", &c[i], &y[i]);        }        for(i=0; i<n-1; i++){            c[i+1]=min(c[i+1],c[i]+s);        }        for(i=0; i<n; i++){            sum+=c[i]*y[i];        }        printf("%lld\n", sum);    }    return 0;}

由于每周的花费一直波动，但仓库的价格不变，所以只须拿下一周的花费与这周的比，选出下一周的最优解。。。依次类推直到结束。。。