POJ 1260 Pearls（DP）

Pearls

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 7545 Accepted: 3733

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

22100 1100 231 101 11100 12

Sample Output

3301344

有 n (1<=n<=100) 个不同档次的珍珠，价格随着档次升高而升高，为了防止顾客只买1 个珍珠，现规定顾客必须加付相当于所买档次10个珍珠的钱。我们发现，有时候将等级低的珍珠用等级高的珍珠来替代购买，反而能省钱，因为此时每个珍珠因为档次升高而多付的钱少于本来需要加付的钱。问，给定各个等级需要买的珍珠数目和该等级每颗珍珠的价钱，怎样的替代方案才能最省钱？

对于一种珍珠（数量: ai 价格:pi），单独购买的花费为 (ai + 10 ) * pi  ;  还可以与下一种珍珠一起购买(ai + ai+1 + 10 ) * pi+1  ,此时ai的价格变为pi+1。 

两种购买方式取花费最小的   min(    (ai + 10 ) * pi +  (ai + 10 ) * pi ，( ai + ai+1 + 10) * pi+1。（一次可选择多种连续的珍珠一起购买）

对于样例1    单独购买的花费为 ( 100 +10 ) * 1 + ( 100 + 10) * 2 =330 ; 一起购买的花费为 ( 100 + 100 +10) *2 = 420,所以选择第一种购买方式

dp[i] 表示购买第i种珍珠的最小花费，初始化的时候为 dp[i] = (a[i]+10) * pi  即每种珍珠都是单独购买的。

不管花费多少，购买的珍珠数量是一定的，s[i] 表示 从第1种到第i种珍珠 一共要买的数量

我们可以通过 dp[i] = min(    (s[i] - s[j] + 10) * p[i] + dp[j]  , dp[i] )  来找dp[i]的最小值，其中(s[i] - s[j] + 10) * pi ,表示从第j种珍珠到第i种珍珠都以pi的价格一起购买

#include <iostream>#include <cstdio>#include <cstring>#include <stdlib.h>using namespace std;const int N = 110;int main(){    int T;    scanf("%d",&T);    while(T--)    {        int n;        int dp[N];        int a[N],p[N];        int s[N];        scanf("%d",&n);        memset(dp,0,sizeof(dp));        memset(s,0,sizeof(s));        for(int i=1; i<=n; i++)        {            scanf("%d%d",&a[i],&p[i]);            s[i]+=s[i-1]+a[i];            dp[i]=dp[i-1]+(a[i]+10)*p[i];        }        for(int i=1; i<=n; i++)            for(int j=0; j<i; j++)                dp[i]=min(dp[i],(s[i]-s[j]+10)*p[i] + dp[j]);        printf("%d\n",dp[n]);    }    return 0;}