POJ2393奶酪工厂

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

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

 

******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[10005],w[10005],s,k = 1;
long long int ans = 0;
int main()
{
    scanf("%d %d",&n,&s);
    for(i = 1;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 0;
}