Yogurt factory
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 16669 | Accepted: 8176 |
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.
看了几篇博客才有了点思路, 每天生产的酸奶是那一天的,要比较今天的 C 与 Cmin + S 的关系, Cmin是前几个周存储在仓库里的酸奶最小成本, 随着时间的推移,Cin+S , 直到他不再小于当前的C, 替换掉。
就是一直再权衡当前C与Cmin+S, 求得最小成本。
#include <iostream> #include <stdio.h> using namespace std; int main() { long long ans; int n, s, y, c, minc; while(scanf("%d%d", &n, &s)!=-1) { ans=0; minc=5005; for(int i=0; i<n; i++) { scanf("%d%d", &c, &y); if(c > minc +s) c = minc + s; minc = c; ans += c*y; } printf("%lld\n", ans); } return 0; }
原文地址:https://www.cnblogs.com/Dawn-bin/p/10805606.html