[UVA][dp] 11400 - Lighting System Design
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 |
778 |
Problemsetter: Mohammad Mahmudur Rahman
Special Thanks to: Manzurur Rahman Khan
題目描述:
要設計會議的照明系統,每種需求燈泡擁有各自的電壓跟花費,而同時會有專屬的電源供應器來供應相同總類的所有燈泡,設計時,已經指派各種燈泡的指定個數,但是可以使用電壓大的代替電壓小的。
求最少花費。
題目解法:
不仿大膽猜測一下,肯定跟電壓大小順序 DP 有關。
再證明幾點
1) 如果要替換某個燈泡時,肯定一次就會將該種燈泡全部替換。不會使用部分替換來減少金額。易想而知,部分替換絕對沒有比全部替換來得好,全部替換還可以省下電源供應器的錢。
-> 事實上還是要看一下函數圖來證明,想必一定是單調的。
2) 替換時,一定是替換前面的連續段。// 假使已經根據電壓大小由小排到大。
-> 不會跳號替換,因為假使跳號時,則表示被忽略的那個無法減少花費,而它的前一個卻能減少花費,可以推估這個解必然也要納入替換中,這裡有點 greedy 的意思在。
最後 dp[i] 表示考慮前 i 個,並且以 i 為最佳解的最後一組型式的最少花費。
O(n^2)
#include <stdio.h>
#include <algorithm>
using namespace std;
struct Light {
int V, K, C, L;
bool operator<(const Light &a) const {
return V < a.V;
}
};
int main() {
int n, i, j;
Light D[1005];
while(scanf("%d", &n) == 1 && n) {
for(i = 0; i < n; i++)
scanf("%d %d %d %d", &D[i].V, &D[i].K, &D[i].C, &D[i].L);
sort(D, D+n);
int dp[1005];
for(i = 0; i < n; i++) {
int sum = 0;
dp[i] = 0xfffffff;
for(j = i; j >= 0; j--) {
sum += D[j].L;
if(j)
dp[i] = min(dp[i], dp[j-1] + D[i].K + sum*D[i].C);
else
dp[i] = min(dp[i], D[i].K + sum*D[i].C);
}
}
printf("%d\n", dp[n-1]);
}
return 0;
}