E - Pearls（dp）

╰+哭是因爲堅強的太久メ 2022-04-24 00:46 126阅读 0赞

### 题目描述： ###

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  
题意：  
给你n个物品需要的价值和数量，请你求出最少花多少钱。注意，价值低的物品可以用价值高的物品代替。并如果买这种物品，则需要额外支付10个数量的这个物品的钱。  
分析:  
一开始我用了一个类似贪心的算法。 但是很遗憾只能过样例。  
我们可以想到，价值小的物品可以用价值高的物品代替。那么我们肯定是价值从小到大dp。先买价值小的物品。在买价值大的物品。然后考虑，在我之前所买的那些物品是否用这个价值高的物品代替会更节约钱。

#include"stdio.h"
    #include"string.h"
    #include"algorithm"
      using namespace std;
    int main()
    {
        int T;
        int n;
        scanf("%d",&T);
        while(T--)
        {
            scanf("%d",&n);
            int cnt[110],p[110];
            int sum[110];
            sum[0]=0;
            cnt[0]=0;p[0]=0;
            for(int i=1;i<=n;i++)
                {
                    scanf("%d%d",&cnt[i],&p[i]);
                    sum[i]=sum[i-1]+cnt[i];//后面表示数量方便
                }
            int dp[110];
            memset(dp,0,sizeof(dp));
            int s=0;
            for(int i=1;i<=n;i++)
                {
                    dp[i]=(cnt[i]+10)*p[i]+dp[i-1];
                    for(int j=0;j<i;j++)
                    {
                        int cont=sum[i]-sum[j]+10;
                        dp[i]=min(dp[i],cont*p[i]+dp[j]);
                        //printf("%d\n",dp[i]);
                    }
                   //8printf("%d\n",dp[i]);
                }
            printf("%d\n",dp[n]);
        }
    }

[family.In]: http://family.In
[needed.No]: http://needed.No