【线性结构上的动态规划】UVa 11400 - Lighting System Design

标签：

Problem F
Lighting System Design

Input: Standard Input

Output: Standard Output

You are given the task to design a lighting system for a huge conference hall. After doing a lot of calculation & 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 & 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.) & complete the design. 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 & 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.

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 & 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.

Output

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

Sample Input                                                  Output for Sample Input

3

100 500 10 20

120 600 8 16

220 400 7 18

0

读题时题意理解的不太好，看分析后才明白。给出n种电灯泡，含四种属性（V-电压,K-电源费用,C-每个灯泡费用,L-所需灯泡数量）；输出最合理的照明系统设计方案，要求花费最少的钱。

自己没考虑到的是：每种电压的灯泡要么全换，要么不换。因为如果只将部分灯泡换成另一种灯泡，则需要买两种不同电压的电源。这样不划算。而且，电压高的灯泡，因电流大小相等，故功率也大，这样会节省更多的钱。

代码很简单：

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstdlib>
 4 #include<cstring>
 5 #include<algorithm>
 6 using namespace std;
 7 const int maxn = 1005;
 8 struct Lamp
 9 {
10     int V, K, C, L;
11     bool operator < (const Lamp& a) const
12     {
13         return V < a.V;
14     }
15 }lamp[maxn];
16 int dp[maxn], sum[maxn];
17 int main()
18 {
19     int n;
20     while(~scanf("%d", &n) && n)
21     {
22         for(int i = 1; i <= n; i++)
23         {
24             scanf("%d%d%d%d", &lamp[i].V, &lamp[i].K, &lamp[i].C, &lamp[i].L);
25         }
26         sort(lamp+1, lamp+1+n);
27         memset(sum, 0, sizeof(sum));
28         memset(dp, 0, sizeof(dp));
29         sum[0] = 0;
30         for(int i = 1; i <= n; i++)
31         {
32             sum[i] = sum[i-1]+lamp[i].L;
33             if(i == 1) dp[i] = lamp[i].C*sum[i]+lamp[i].K;
34             else
35             for(int j = 0; j < i; j++)
36             {
37                 if(dp[i] == 0) dp[i] = dp[j]+lamp[i].C*(sum[i]-sum[j])+lamp[i].K;
38                 else dp[i] = min(dp[j]+lamp[i].C*(sum[i]-sum[j])+lamp[i].K, dp[i]);
39             }
40         }
41         printf("%d\n", dp[n]);
42     }
43     return 0;
44 }

【线性结构上的动态规划】UVa 11400 - Lighting System Design

标签：

原文地址：http://www.cnblogs.com/LLGemini/p/4312195.html