SDAU 贪心专题 15 牛奶问题

最新推荐文章于 2021-02-15 13:13:09 发布

Lee丶0767

最新推荐文章于 2021-02-15 13:13:09 发布

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分类专栏：贪心算法文章标签： acm

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1：问题描述
Problem 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

2：大致题意

先分别输入天数和存一天牛奶需要花费的钱数。牛奶只能存一天。
下面输入的是今天牛奶的单价和需求量。
求出花费最少的钱数。
3：思路

从第2天开始，每次买牛奶之前做一次判断，比较今天的单价和（昨天的单价+5）。由此判断出是今天做牛奶还是用昨天的牛奶。
由于牛奶只能存放一天这个条件，这个题简单了许多哎。

4：感想

懂了题意就很简单的一道题。但是。。。
英语是硬伤啊〒_〒

5：ac代码

#include <cstdio>
#include<iostream>
#include<stdio.h>
#include<vector>
#include<algorithm>
#include<numeric>
#include<math.h>
#include<string.h>
#include<map>
#include<set>
#include<vector>
#include<iomanip>
using namespace std;
int main()
{
    int n,m,i,j,z1,z2,k;
    long long sum;
    int a[100000],b[100000];
    cin>>n>>m;
    for(i=0;i<n;i++)
    {
        cin>>a[i]>>b[i];
    }
    sum=0;
    for(k=0;k<n;k++)
    {
        if(k==0) sum+=a[k]*b[k];
        else
        {
            z1=a[k-1]*b[k]+m*b[k];
            z2=a[k]*b[k];
            if(z1>z2) sum+=z2;
            else sum+=z1;
        }
    }
    cout<<sum<<endl;
    return 0;
}