TOJ 2888 Pearls

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

2
2
100 1
100 2
3
1 10
1 11
100 12

Sample Output

330
1344

Source

Northwestern Europe 2002

不应该排序的，排完序就错了。

dp[i]：存放购买前i个Pearls的最优解。

则有dp[i]=min(dp[j]+(s[i]-s[j]+10)*p[i],dp[i])

#include <stdio.h>
#include <iostream>
using namespace std;
int main()
{
    int t,c;
    int a[];
    int p[];
    int dp[];
    scanf("%d",&t);
    while(t--){
        int s[]={};
        scanf("%d",&c);
        for(int i=; i<c; i++){
            scanf("%d %d" ,&a[i] ,&p[i]);
            if(i==){
                s[i]=a[i];
            }else{
                s[i]=s[i-]+a[i];
            }
            dp[i]=(s[i]+)*p[i];
        }
        for(int i=; i<c; i++){
            for(int j=; j<=i; j++){
                dp[i]=min( dp[j]+(s[i]-s[j]+)*p[i] , dp[i] );
            }
        }
        printf("%d\n",dp[c-]);
    }
    return ;
}