POJ 2393 Yogurt factory__贪心

最新推荐文章于 2020-05-12 10:40:11 发布

原创最新推荐文章于 2020-05-12 10:40:11 发布 · 481 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#poj #算法基础 #贪心

数据结构与算法同时被 2 个专栏收录

41 篇文章

订阅专栏

重拾算法之路-Coursera算法基础

40 篇文章

订阅专栏

本文介绍了一群奶牛开设酸奶厂的案例，通过贪心算法计算出在满足每周酸奶需求的同时，如何达到最低的成本消耗。文章提供了完整的代码实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

分析

题目大意：
一群奶牛建了个酸奶厂，要在N周内每周交Yi单位的酸奶，每单位酸奶生产单位成本为Ci,存储单价为S，不用担心生产量限制与酸奶存储过期问题，求按计划交付的最小成本。

思路：
用贪心算法解决，求成本最小值，每周用最小单位成本交付，最小单位成本为这周生产单位成本或上周判定的最小单位成本加一周的存储费用。

实现

#include <iostream>

int main()
{
//  freopen("in.txt", "r", stdin);
    int n, s, c, y, minC = 5000;
    long long total = 0;
    scanf("%d%d", &n, &s);
    for(int i = 0; i < n; i++) {
        scanf("%d%d", &c, &y);
        minC = (c < (minC + s)) ? c : (minC + s);
        total += minC * y;
    }
    printf("%lld\n", total);
    return 0;
}