poj 1260 dp

时间：2016-08-01 19:04:31 阅读：167 评论：0 收藏：0 [点我收藏+]

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Description

In Pearlania everybody is fond of pearls. One company, called The Royal Pearl, produces a lot of jewelry with pearls in it. The Royal Pearl has its name because it delivers to the royal family of Pearlania. But it also produces bracelets and necklaces for ordinary people. Of course the quality of the pearls for these people is much lower then the quality of pearls for the royal family.In Pearlania pearls are separated into 100 different quality classes. A quality class is identified by the price for one single pearl in that quality class. This price is unique for that quality class and the price is always higher then the price for a pearl in a lower quality class.
Every month the stock manager of The Royal Pearl prepares a list with the number of pearls needed in each quality class. The pearls are bought on the local pearl market. Each quality class has its own price per pearl, but for every complete deal in a certain quality class one has to pay an extra amount of money equal to ten pearls in that class. This is to prevent tourists from buying just one pearl.
Also The Royal Pearl is suffering from the slow-down of the global economy. Therefore the company needs to be more efficient. The CFO (chief financial officer) has discovered that he can sometimes save money by buying pearls in a higher quality class than is actually needed.No customer will blame The Royal Pearl for putting better pearls in the bracelets, as long as the
prices remain the same.
For example 5 pearls are needed in the 10 Euro category and 100 pearls are needed in the 20 Euro category. That will normally cost: (5+10)*10+(100+10)*20 = 2350 Euro.Buying all 105 pearls in the 20 Euro category only costs: (5+100+10)*20 = 2300 Euro.
The problem is that it requires a lot of computing work before the CFO knows how many pearls can best be bought in a higher quality class. You are asked to help The Royal Pearl with a computer program.

Given a list with the number of pearls and the price per pearl in different quality classes, give the lowest possible price needed to buy everything on the list. Pearls can be bought in the requested,or in a higher quality class, but not in a lower one.

Input

The first line of the input contains the number of test cases. Each test case starts with a line containing the number of categories c (1<=c<=100). Then, c lines follow, each with two numbers ai and pi. The first of these numbers is the number of pearls ai needed in a class (1 <= ai <= 1000).
The second number is the price per pearl pi in that class (1 <= pi <= 1000). The qualities of the classes (and so the prices) are given in ascending order. All numbers in the input are integers.

Output

For each test case a single line containing a single number: the lowest possible price needed to buy everything on the list.

Sample Input

Sample Output

330
1344

题意：有c种品质不同的珠宝，若要买某一品质的珠宝必须支付(ai+10)*pi，可以用高品质的珠宝代替低品质的

(通过减少购买的种类来节约多支付的10个的价钱）.求要买到所有目标数量的珠宝至少要花多少钱

题解：注意题目算价格的方式注意只能由高品质代替低品质

注意题目中 given in ascending order 珠宝的品质是升序给出的

dp[i] 表示以第i个品质结尾的最优解

dp[i]=min{dp[j-1]+(sum[i]-sum[j-1]+10)*p[i]}

(sum[i]-sum[j-1]+10)*p[i]表示第j种到第i种珍珠全部替换为第i种

 1 /******************************
 2 code by drizzle
 3 blog: www.cnblogs.com/hsd-/
 4 ^ ^    ^ ^
 5  O      O
 6 ******************************/
 7 //#include<bits/stdc++.h>
 8 #include<iostream>
 9 #include<cstring>
10 #include<cstdio>
11 #include<map>
12 #include<algorithm>
13 #include<cmath>
14 #define ll long long
15 #define PI acos(-1.0)
16 #define mod 1000000007
17 using namespace std;
18 int t;
19 int n;
20 int cost[105];
21 int qu[105];
22 int sum[105];
23 int dp[105];
24 int main()
25 {
26     scanf("%d",&t);
27     for(int i=1; i<=t; i++)
28     {
29         scanf("%d",&n);
30         sum[0]=0;
31         for(int j=1; j<=n; j++)
32         {
33             scanf("%d %d",&cost[j],&qu[j]);
34             sum[j]=sum[j-1]+cost[j];
35             dp[j]=1e9;
36         }
37         dp[0]=0;
38         for(int j=1; j<=n; j++)
39         {
40             for(int k=1; k<=j; k++)
41                 dp[j]=min(dp[j],dp[k-1]+(sum[j]-sum[k-1]+10)*qu[j]);
42         }
43         printf("%d\n",dp[n]);
44     }
45     return 0;
46 }

poj 1260 dp

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原文地址：http://www.cnblogs.com/hsd-/p/5726621.html

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