uva 11400 Lighting System Design （dp）

题目链接：uva 11400 - 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

 

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Problemsetter: Mohammad Mahmudur Rahman

Special Thanks to: Manzurur Rahman Khan

 

题目大意：

多组测试数据 0 结束，输入灯的种类数 n ， 然后 n 行每个依次是灯需要的电压，购买这种电压的钱，这种电压下的每盏灯的钱，灯的数目。

条件：低电压的灯可以用高电压代替。

输出：购买这些灯需要的最小的钱数。

首先种类按电压从高到低排序，高电压去更新低电压的花费

用 dp[k] 记录装到第k种需要的最小的花费，维护更新这个数组即可

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;

const int maxn=1010;
struct node{
    int v,cost,per,cnt;
    friend bool operator < (node a,node b){
        return a.v>b.v;
    }
}a[maxn];
int n,dp[maxn];

void input(){
    for(int i=0;i<n;i++){
        scanf("%d%d%d%d",&a[i].v,&a[i].cost,&a[i].per,&a[i].cnt);
    }
}

void computing(){
    sort(a,a+n);
    dp[0]=a[0].cost+a[0].per*a[0].cnt;
    for(int i=1;i<n;i++) dp[i]=dp[i-1]+a[i].cnt*a[0].per;
    for(int i=1;i<n;i++){
        int sum=a[i].cost+dp[i-1];
        for(int j=i;j<n;j++){
            sum+=a[j].cnt*a[i].per;
            if(sum<dp[j]) dp[j]=sum;
        }
    }
    printf("%d\n",dp[n-1]);
}

int main(){
    while(scanf("%d",&n)!=EOF && n>0){
        input();
        computing();
    }
    return 0;
}

转载于:https://www.cnblogs.com/toyking/p/3797354.html