Lighting System Design - UVa 11400 dp

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

778

題意：每種類型的燈有不同的電壓，且隻需要買一個電源，你可以将其中某些燈泡換成更大功率的（也就是更大電壓的），問最少的花費是多少。

思路：将電壓從小到大排序，分别編号1-n種，可以想象一下，最後的結果肯定是1到m1種的燈泡都換成m1的，m1+1到m2種的燈泡都換成m2的，那麼我們dp[i]表示成買了第i個的電源，并且解決了前i種燈泡的最少花費。

AC代碼如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct node
{ int v,k,c,l;
}lap[1010];
bool cmp(node a,node b)
{ return a.v<b.v;
}
int n,sum[1001],dp[1010],INF=1000000000;
int main()
{ int i,j,k;
  while(~scanf("%d",&n) && n)
  { for(i=1;i<=n;i++)
     scanf("%d%d%d%d",&lap[i].v,&lap[i].k,&lap[i].c,&lap[i].l);
    sort(lap+1,lap+1+n,cmp);
    for(i=1;i<=n;i++)
     sum[i]=sum[i-1]+lap[i].l;
    for(i=1;i<=n;i++)
    { dp[i]=INF;
      for(j=i-1;j>=0;j--)
       dp[i]=min(dp[i],dp[j]+lap[i].k+lap[i].c*(sum[i]-sum[j]));
    }
    printf("%d\n",dp[n]);
  }
}