hdu 1300 Pearls

Pearls

Time Limit: 2000/1000 MS(Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2980    Accepted Submission(s): 1503

Problem Description

In Pearlaniaeverybody is fond of pearls. One company, called The Royal Pearl, produces alot of jewelry with pearls in it. The Royal Pearl has its name because itdelivers to the royal family of Pearlania. But it also produces bracelets andnecklaces for ordinary people. Of course the quality of the pearls for thesepeople is much lower then the quality of pearls for the royal family. InPearlania pearls are separated into 100 different quality classes. A qualityclass 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 thenthe price for a pearl in a lower quality class.

Every month the stock manager of The Royal Pearl prepares a list with thenumber of pearls needed in each quality class. The pearls are bought on thelocal pearl market. Each quality class has its own price per pearl, but forevery complete deal in a certain quality class one has to pay an extra amountof money equal to ten pearls in that class. This is to prevent tourists frombuying 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 financialofficer) has discovered that he can sometimes save money by buying pearls in ahigher quality class than is actually needed. No customer will blame The RoyalPearl for putting better pearls in the bracelets, as long as the prices remainthe same.

For example 5 pearls are needed in the 10 Euro category and 100 pearls areneeded 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 = 2300Euro.

The problem is that it requires a lot of computing work before the CFO knowshow many pearls can best be bought in a higher quality class. You are asked tohelp The Royal Pearl with a computer program.

Given a list with the number of pearls and the price per pearl in differentquality classes, give the lowest possible price needed to buy everything on thelist. Pearls can be bought in the requested, or in a higher quality class, butnot in a lower one.

 

 

Input

The first line ofthe input contains the number of test cases. Each test case starts with a linecontaining the number of categories c (1 <= c <= 100). Then, c linesfollow, each with two numbers ai and pi. The first of these numbers is thenumber of pearls ai needed in a class (1 <= ai <= 1000). The secondnumber is the price per pearl pi in that class (1 <= pi <= 1000). Thequalities of the classes (and so the prices) are given in ascending order. Allnumbers in the input are integers.

 

 

Output

For each test casea single line containing a single number: the lowest possible price needed tobuy everything on the list.

 

 

Sample Input

2

2

100 1

100 2

3

1 10

1 11

100 12

 

 

Sample Output

330

1344

题目链接：点击打开链接

题意：

现需要买n类珍珠，等级按升序方式给出每一类需要的个数和价格，价格低的可以由价格高的代购，每一次购买需多购买10颗，求最小花费。高级的可以代替低级的。

题解

用dp[i]表示买前i种珍珠所需要的最少的钱数
根据贪心，买第i种珍珠时，买的越多越划算。
也即，对于第i种珍珠，要么一个都不买，要么把前面更便宜的没买的和本组该买的都买了

因此，对于每一层， 只有买和不买两种选择
并且不可能出现便宜的没买但是贵的买了

dp[i]= min{ dp[i] , dp[j] + (sum[i] -sum[j] + 10)*price[i] } 1<=i<=n,0<=j<i

 

#include <bits/stdc++.h>

using namespace std;
#define N 3000
#define INF 0x3f3f3f3f
struct node {
    int cost;
    int number;
}a[N];
int dp[N];
int sum[N];
int main()
{
    int T;
    scanf("%d", &T);
    while(T--) {
        int n;
        scanf("%d", &n);
        for(int i = 1; i <= n; i++)
            scanf("%d%d", &a[i].number, &a[i].cost);
        for(int i = 1; i <= n; i++)
            sum[i] = sum[i-1] + a[i].number;
        memset(dp, INF, sizeof(dp));
        dp[0] = 0;
        dp[1] = (a[1].number+10)*a[1].cost;
        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]+10)*a[i].cost);
            }
        }
        printf("%d\n", dp[n]);
    }
    return 0;
}