bzoj 16801740 : [Usaco2005 Mar]Yogurt factory 贪心 双倍经验

1680: [Usaco2005 Mar]Yogurt factory

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.

吨酸奶。酸奶的生产受到很多因素的影响，所以每个月的生产成本是变化的，其中第 i 个月的成本是

每吨 Ci 元。

奶牛可以提前里把酸奶做好，存在仓库里，等需要的时候再拿出来卖。存储在仓库里的酸奶，每

吨酸奶存放一个月需要支付 S 元的维护费用，存放的时间可以任意长。假设工厂的产量是无限的，存

储酸奶的仓库也是无限大的。请问为了满足订单的需要，奶牛生产这些酸奶最少要花多少钱？

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.

• 第一行：两个整数 N 和 S，1 ≤ N ≤ 10000, 1 ≤ S ≤ 100

• 第二行到第 N + 1 行：第 i + 1 行有两个整数 Ci 和 Ai，1 ≤ Ci ≤ 5000, 1 ≤ Ai ≤ 10000

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.

• 单个整数：表示生产酸奶的最小总费用

Sample Input

4 5
88 200
89 400
97 300
91 500

Sample Output

126900

OUTPUT DETAILS:
第一个月生产 200 吨酸奶；第二个月生产
700 吨酸奶，并存下 300 吨；第三个月不生产酸
奶；第三个月生产 500 吨

思路 ：

　　每天选择最小价格做酸奶， 最小价格是之前的价格加上相应的储存价格和今天的价格取min

代码 ：

 

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
int tmp;
long long ans;
struct node
{
    long long a,c;
}d[10001];
int main()
{
    /*freopen("factory.in","r",stdin);
    freopen("factory.out","w",stdout);*/
    int n,s;
    scanf("%d%d",&n,&s);
    scanf("%lld%lld",&d[1].c,&d[1].a);
    tmp=d[1].c;
    ans+=tmp*d[1].a;
    for(int i=2;i<=n;i++)    
    {
        tmp+=s;
        scanf("%lld%lld",&d[i].c,&d[i].a);
        if(tmp>=d[i].c)
            tmp=d[i].c;
        ans+=tmp*d[i].a;
    }
    printf("%lld",ans);
}