- 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. Yuckys 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 Yogurts 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 Yuckys 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.
题目大意:
牛们收购了一个奶酪工厂, 接下来的 N 个星期里, 牛奶价格和劳力价格不断起伏.第 i 周, 生产一个单位奶酪需要 Ci(1Ci5000) 便士.工厂有一个货栈, 保存一单位奶酪, 每周需要 S(1S100) 便士, 这个费用不会变化.货栈十分强大, 可以存无限量的奶酪, 而且保证它们不变质.工厂接到订单, 在第 i 周需要交付 Yi(0Yi104) 单位的奶酪给委托人.第 i 周刚生产的奶酪, 以及之前的存货, 都可以作为产品交付.请帮牛们计算这段时间里完成任务的最小代价.
第 1 行输入两个整数 N 和 S.接下来 N 行输入 Ci 和 Yi.
输出最少的代价.注意, 可能超过 32 位长整型
提示翻译:
第 1 周生产 200 单位奶酪并全部交付; 第 2 周生产 700 单位, 交付 400 单位, 有 300 单位; 第 3 周交
付 300 单位存货.第 4 周生产并交付 500 单位.
思路: 更新下一周的生产成本即可
代码如下:
- #include <iostream>
- #include <cstdio>
- using namespace std;
- const int maxn = 10005;
- struct node
- {
- int c,y;
- };
- node a[maxn];
- int main()
- {
- int n,s;
- long long sum;
- while(~scanf("%d%d",&n,&s))
- {
- for(int i=0;i<n;i++)
- {
- scanf("%d%d",&a[i].c,&a[i].y);
- }
- sum=0;
- for(int i=0;i<n;i++)
- {
- sum+=a[i].c*a[i].y;
- if(i!=n-1);
- {
- a[i+1].c=min(a[i+1].c,a[i].c+s);
- }
- }
- cout << sum << endl;
- }
- return 0;
- }
- View Code
来源: http://www.bubuko.com/infodetail-2501202.html