uva11400 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

 题意：有n种照明系统，每种照明系统要L个灯泡，每个灯泡花费C，问最少花费多少组成这个照明系统。

高电压可以代替低电压，所以枚举每一组电压，dp[i]是表示用第i种电源组成系统花费的最少费用，转移方程是dp[i]=min(dp[i],dp[j-1]+p[j].num*p[i].ncost+p[i].vcost)。里面num是代表这套系统要运行灯泡要多少个,i是代表用这个电压灯泡的费用，vcost是拥有这个电压的费用，dp【n】就是答案了。

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int MAXN=1010;const int INF=99999999;int dp[MAXN];struct node{    int v,vcost,ncost,num;}p[MAXN];bool cmp(node x,node y){    return x.v<y.v;}int main(){    int n,i;    while(scanf("%d",&n)!=EOF&&n)    {        for(i=1;i<=n;i++)        {            scanf("%d%d%d%d",&p[i].v,&p[i].vcost,&p[i].ncost,&p[i].num);        }        sort(p+1,p+n+1,cmp);        for(i=1;i<=n;i++)            dp[i]=INF;        dp[0]=0;        int sum;        for(i=1;i<=n;i++)        {            sum=p[i].vcost;            for(int j=i;j>=1;j--)            {                sum+=p[j].num*p[i].ncost;                dp[i]=min(dp[i],dp[j-1]+sum);            }        }        printf("%d\n",dp[n]);    }    return 0;}