UVA11400 Lighting System Design（DP）

You are given the task to design a lighting system for a huge conference hall. After doing a lot of calculation and 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 and 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.) and 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 and 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 and 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
3

100 500 10 20

120 600 8 16

220 400 7 18

0
Sample Output
778

题意：

就是说先给你一个n，代表这个系统中有n中类型的灯泡，v是这种类型灯泡的电压，每种电压的灯泡有需要一个变压器吧（我自己理解的），不管这种类型的灯泡有多少个，只需要一个这种电压变压器就可以了，第三个值是安装一个这样的灯泡多少钱，第四个值是目前这个系统中有这种灯泡多少个

我们可以用电压大的灯泡去代替电压小的灯泡，求最后这套系统的最小花费

题解：

那肯定如果某种灯泡要换那么肯定是一下子把这种灯泡一下换完（想想就知道了），那么我们对数据先初始化，对其电压从小到大排序，之后再求前i中灯泡的前缀和

每种灯泡要么换要么不换

设sum[i]为前i种灯泡的总数量（即L值之和），d[i]为灯泡1～i的最小花费，

动态转移方程就是d[i]=min(d[j]+(sum[i]-sum[j])*c[i]+k[i],d[i])；

代码：

//要先排序，后求和<_>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
typedef long long ll;
const int maxn=1005;
const int INF=0x3f3f3f3f;
int dp[maxn],v[maxn],ans[maxn];
struct shudui
{
    int v,k,c,l;
} m[maxn];
bool mmp(shudui x,shudui y)
{
    return x.v<y.v;
}
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        if(n==0) break;
        int sum=0;
        for(int i=1; i<=n; ++i)
            scanf("%d%d%d%d",&m[i].v,&m[i].k,&m[i].c,&m[i].l);
            //ans[i]=m[i].k+m[i].c*m[i].l;
        sort(m+1,m+1+n,mmp);
        for(int i=1;i<=n;++i)
            v[i]=v[i-1]+m[i].l;
        memset(dp,INF,sizeof(dp));
        dp[0]=0;
        for(int i=1;i<=n;++i)
        {
            for(int j=0;j<i;++j)
            {
                dp[i]=min(dp[i],dp[j]+(v[i]-v[j])*m[i].c+m[i].k);
            }
        }
        printf("%d
",dp[n]);
    }
    return 0;
}