UV 11400 Lighting System Design (DP lighting system design)

To design a lighting system in a certain place, a total of n Different Types of bulbs are required. Then input the voltage of each bulbs. V corresponds to the price of the voltage power supply. K. The price of each bulb. C. The quantity of such bulbs. l voltage. low bulbs can be replaced with high-voltage bulbs. Each type of bulbs only requires a corresponding power supply to achieve the minimum cost of the lighting system.

Relatively simple dp it is easy to know that to replace one of the bulbs with a higher voltage, it is only possible to reduce the total cost by replacing all the bulbs with the other. this saves the power cost of this light bulb. First, sort the light bulb by voltage. Then, each light bulb can replace any light bulb in front of it. Then, s [I] indicates the total number of the first I bulbs. s [I]-s [J-1] indicates the total number of bulbs from type J to type [I]

If d [I] represents the minimum cost of the first I bulb, then the equation d [I] = min {d [J-1] + (s [I]-s [J-1]) can be obtained. * C [I] + k [I]} J is all numbers between 1 and I

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int N = 1005;int d[N], s[N], v[N], k[N], c[N], l[N], o[N], n;bool cmp (int i, int j){    return v[i] < v[j];}int main(){    while (~scanf ("%d", &n), n)    {        for (int i = 1; i <= n; ++i)        {            scanf ("%d%d%d%d", &v[i], &k[i], &c[i], &l[i]);            o[i] = i;        }        sort (o + 1, o + n + 1, cmp);        memset (d, 0x3f, sizeof (d));        d[0] = 0;        for (int i = 1; i <= n; ++i)        {            s[i] = s[i - 1] + l[o[i]];            for (int j = 1; j <= i; ++j)                d[i] = min (d[i], d[j - 1] + (s[i] - s[j - 1]) * c[o[i]] + k[o[i]]);        }        printf ("%d\n", d[n]);    }    return 0;}<span style="font-family:Comic Sans MS;font-size:12px;"></span><span style="font-size:12px;"></span>

 

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 lampsNDifferent 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 shoshould 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

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UV 11400 Lighting System Design (DP lighting system design)