POJ2393奶酪工厂

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Yogurt factory

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 14771		Accepted: 7437

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

Sample Output

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

******每次判断哪个地方更适合做单个奶酪，就将从现在开始及以后都按这套方案走，如果遇到更优的方案当然就修改方案啦，这就是传说中的贪心啦。

 #include<cstdio>

 #include<cstring>

 #include<algorithm>

 #include<cmath>

 using namespace std;

 int i,j,n,c[],w[],s,k = ;

 long long int ans = ;

 int main()

 {

     scanf("%d %d",&n,&s);

     for(i = ;i <= n;i++)

     {

         scanf("%d %d",&c[i],&w[i]);

         if(c[i] <= c[k] + (i - k) * s)

         k = i;

         ans += c[k] * w[i] + (i - k) * s * w[i];

     }

     printf("%lld",ans);

     return ;

 }