UVA11400-Lighting System Design（动态规划基础）

Problem UVA11400-Lighting System Design

Accept: 654  Submit: 4654
Time Limit: 3000 mSec

Problem Description

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

题解：dp[i]表示前i种灯的最小花费并且i这种灯的电源被使用了，那么状态转移就很简单了，就是前j种（j<i）灯的最小花费，加上从j+1到i种灯的最小花费。

#include <bits/stdc++.h>

using namespace std;

const int maxn = 1000 + 10;
const int INF = 0x3f3f3f3f;

int n;
int sum[maxn], dp[maxn];

struct Lamp {
    int v, k, c, l;
    bool operator < (const Lamp &a)const {
        return v < a.v;
    }
}lamp[maxn];

int main()
{
    //freopen("input.txt", "r", stdin);
    while (~scanf("%d", &n) && n) {
        for (int i = 1; i <= n; i++) {
            scanf("%d%d%d%d", &lamp[i].v, &lamp[i].k, &lamp[i].c, &lamp[i].l);
        }
        sort(lamp + 1, lamp + n + 1);

        memset(dp, INF, sizeof(dp));
        sum[0] = dp[0] = 0;
        for (int i = 1; i <= n; i++) {
            sum[i] = sum[i - 1] + lamp[i].l;
        }
        for (int i = 1; i <= n; i++) {
            for (int j = 0; j < i; j++) {
                dp[i] = min(dp[i], dp[j] + (sum[i] - sum[j])*lamp[i].c + lamp[i].k);
            }
        }

        printf("%d
", dp[n]);
    }
    return 0;
}