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 0

778

题意：每种类型的灯有不同的电压，且只需要买一个电源，你可以将其中某些灯泡换成更大功率的（也就是更大电压的），问最少的花费是多少。

思路：将电压从小到大排序，分别编号1-n种，可以想象一下，最后的结果肯定是1到m1种的灯泡都换成m1的，m1+1到m2种的灯泡都换成m2的，那么我们dp[i]表示成买了第i个的电源，并且解决了前i种灯泡的最少花费。

AC代码如下：

#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
);
  }
}