Hduoj1300【DP】

/*PearlsTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 1704    Accepted Submission(s): 782Problem DescriptionIn 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.InputThe 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.OutputFor each test case a single line containing a single number: the lowest possible price needed to buy everything on the list.Sample Input22100 1100 231 101 11100 12 Sample Output3301344 SourceNorthwestern Europe 2002 RecommendEddy   |   We have carefully selected several similar problems for you:  1227 1224 1158 1080 1422 */#include<stdio.h>#define MAX  100000000int num1[110], money[110], dp[110];int Min(int x,int y){return x<y?x:y;}int main(){int i, j, k, t, n;scanf("%d", &t);while(t--){scanf("%d", &n);dp[0] = 0;for(i = 1; i <= n; ++i){dp[i] = MAX;scanf("%d%d", &num1[i], &money[i]);if(i == 1)dp[i] = (num1[i]+10) * money[i];else{for(j = 0; j < i; ++j)//make every point to be a new start{int sum = 0;for(k = j+1; k < i; ++k)//get the numbersum += num1[k];dp[i] = Min(dp[i], dp[j] + (sum + num1[i] + 10) * money[i]);//count the min}}}printf("%d\n", dp[n]);}return 0;}

题意：给出一些珍珠，每一种珍珠都有一个单价，并且每种珍珠都需要一定的数量，每种珍珠都有一个品级，现在输入时是按照品级从低到高输入的，现在求买完这些珍珠需要多少钱，其中这里每一种品级的珍珠购买时都需要多买10个，由于这个要求现在规定可以将低等级的珍珠合并到高级的珍珠当中去买，求怎样购买这些珍珠才能使总钱数最小化。

思路：这里用的是DP的思想，关键就是转移方程dp【i】 = min（dp【i】， dp【j】 + sum合并），这里dp代表买到第i种珍珠时所需要的最小花费，sum合并代表将j品级后的珍珠全部合并到i品级的数量中一次性购买，由于这里输入都是从低等级到高等级已经排好序的，所以方程适用于题目要求。