UVa 11400 Lighting System Design(DP 照明设计)

意甲冠军  地方照明系统设计  总共需要n不同类型的灯泡  然后进入 每个灯电压v  相应电压电源的价格k  每一个灯泡的价格c   须要这样的灯泡的数量l   电压低的灯泡能够用电压高的灯泡替换   每种灯泡仅仅须要一个相应的电源   求完毕这个照明系统的最少花费

比較简单的DP  easy知道 当要替换一种灯泡中的一个到令一种电压较高的灯泡时  仅仅有所有替换这样的灯泡为还有一种时才可能使总花费变小   所有替换掉就省下了这样的灯泡的电源花费   先把灯泡依照电压排序   那么每种灯泡都能够替换他前面的随意灯泡了    令s[i]表示前i种灯泡的总数   那么s[i]-s[j-1]表示的是第j种到第[i]种灯泡的总数

令d[i]表示前i种灯泡的最少花费  那么能够得到转移方程  d[i]=min{d[j-1]+(s[i]-s[j-1])*c[i]+k[i]}   j为1到i之间全部数

#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;
}

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



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