照明系统设计（动态规划）- UVA 11400

Posted 2021-04-18 ACM算法日常

You are given the task to design a lighting system for a huge conference hall. After doing a lot of calculation and sketching, you have figured out the requirements for an energy-efficient design that can properly illuminate the entire hall. According to your design, you need lamps of n different power ratings. For some strange current regulation method, all the lamps need to be fed with the same amount of current. So, each category of lamp has a corresponding voltage rating. Now, you know the number of lamps and cost of every single unit of lamp for each category. But the problem is, you are to buy equivalent voltage sources for all the lamp categories. You can buy a single voltage source for each category (Each source is capable of supplying to infinite number of lamps of its voltage rating.) and complete the design.

你接到了为会议礼堂设计照明系统的任务，一番精算之后你已设计完毕。在你的设计中，需要n种不同能量级别的灯。而依据某神奇的电流定律，所有灯需要通过等量电流，所以每种灯有一个对应的电压等级。现在你知道每种灯的所需数量和单价，问题是你还得给每种灯买电压源。每种灯有它对应的电压源，一个源可以为无限盏灯供能。

 

But the accounts section of your company soon figures out that they might be able to reduce the total system cost by eliminating some of the voltage sources and replacing the lamps of that category with higher rating lamps. Certainly you can never replace a lamp by a lower rating lamp as some portion of the hall might not be illuminated then. You are more concerned about money-saving than energy-saving. Find the minimum possible cost to design the system.

然鹅财政部门要削减开支……为了省钱，可以把一些灯泡换成电压更高的灯泡，这样就省了好几种电压源的钱，因为只花了一份电压源的钱（看样例可知电压源贼贵）。算出最省钱方案。

Input 

Each case in the input begins with n (1 ≤ n ≤ 1000), denoting the number of categories. Each of the following n lines describes a category. A category is described by 4 integers - V (1 ≤ V ≤ 132000), the voltage rating, K (1 ≤ K ≤ 1000), the cost of a voltage source of this rating, C (1 ≤ C ≤ 10), the cost of a lamp of this rating and L (1 ≤ L ≤ 100), the number of lamps required in this category. The input terminates with a test case where n = 0. This case should not be processed.

每组样例先给出n，代表种类数目。接下来的n行每行将依次给出这种灯的参数——V：电压级别、K：这种电压的电压源价格、C：这种灯泡的单价、L：所需这种灯泡的数量。

输入的n==0时结束输入。

Output
For each test case, print the minimum possible cost to design the system.

对于每组数据，输出最小花费。

Sample Input
3 

100 500 10 20 

120 600 8 16 

220 400 7 18 

0
Sample Output
778

 

样例解释：778 = 400 + （20+16+18）*7

 

只看表层题面不好解决，要进行一定的思考推演出万变情况中不变的规律，方可写出状态转移。

 

结论1：每种灯泡要么不换，要么全换。

如果单拿两种灯泡，我们什么情况换？在我们发现应该换的情况下，如何证明全换肯定比部分换省钱？不妨动笔写写。

结论2：现我们将输入按照电压等级从小到大排序后，j < i，如果 j 要换成 i，则从 j 到 i-1 都要换成 i。

假设 j 要换成 i，而 j+1 是独立不换的，j 换成 i 比换成 j+1更好，那么？

如果结论2成立，那么意味着这个升序列中有几个“断层”，它们之间的弱势灯泡要被替换为右侧大功率灯泡。首先，第n个由于功率最大，肯定是最后一个断层。然后，此时假设断层 i 的上一个断层为 j，由于 j 和 i 之间的灯泡都要换成 i，所以dp[i] = dp[j] + 灯泡数量*i的单价 + i的电压源价格。

该暴力的时候也得暴力一下，上一个断层 j 无法直接得出，可以遍历1~i-1看看哪种方案最优得之。

故给出状态转移方程dp[i] = min{ dp[j] + (s[i]-s[j])*C[i] + K[i] }。其中s[i]是灯泡数量前缀和。

 

注：对于上述结论、思路有疑义或想法的同学欢迎讨论！也欢迎分享其他解法，或者对题目难度进行反馈！

下面是我的代码，欢迎优化、指正：

#include <cstdio>
#include <cstring>
#include <algorithm>

struct category
{
    int V, K, C, L;
}a[1001];

int n;
int dp[1001], s[1001];

int main()
{
    while (~scanf("%d", &n) && n)
    {
        for (int i = 1; i <= n; i++)
            scanf("%d%d%d%d", &a[i].V, &a[i].K, &a[i].C, &a[i].L);

        std::sort(a+1, a+1+n, [](const category& x, const category& y){return x.V < y.V;});

        memset(dp, 0, sizeof(dp));
        for (int i = 1; i <= n; i++)
            s[i] = s[i-1] + a[i].L;

        for (int i = 1; i <= n; i++)
            for (int j = 0; j < i; j++)
                if (!dp[i] || dp[i] > dp[j] + (s[i]-s[j])*a[i].C + a[i].K)
                    dp[i] = dp[j] + (s[i]-s[j])*a[i].C + a[i].K;

        printf("%d\n", dp[n]);
    }
}

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