POJ 1260 简单dp问题 Pearls

PearlsTime Limit: 1000MS      Memory Limit: 10000KTotal Submissions: 9680     Accepted: 4946DescriptionIn 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 12Sample Output3301344

题目大意：给你T组样例，每组样例有n个等级的珍珠。然后有n行，等级呈上升序列输入，每行两个数 需要买的数量p和价格c。买珍珠时有两种购买方法，第一种是分开买，但是没买一种等级的珍珠就会多加10*c（此等级珍珠的价格）。第二种是用等级高的珍珠来替换等级低的珍珠。让你求最小花费。

思路：这道题目明显可以先将dp数组中存上全部买高一等级的珍珠的总花非费。然后在通过和 将其一个一个分开买进行比较，取小值然后更新dp数组，最后输出dp{n}，则就是你要求的结果。

#include <iostream>#include <algorithm>#include <cstdio>#include <string.h>using namespace std;const int N=1e5+10;struct F{    int p,c;} a[N];int dp[N];int sum[N];int main(){    int T;    cin>>T;    while(T--)    {        memset(a,0,sizeof(a));        memset(dp,0,sizeof(dp));        memset(sum,0,sizeof(sum));        sum[0]=0;        int n;        cin>>n;        for(int i=1; i<=n; i++)            cin>>a[i].p>>a[i].c,sum[i]=sum[i-1]+a[i].p,dp[i]=(10+sum[i])*a[i].c;//dp中存的全买一样的花费        for(int i=1;i<=n;i++)            for(int j=i-1;j>=1;j--)        {               dp[i]=min(dp[i],(sum[i]-sum[j]+10)*a[i].c+dp[j]);//和 分开买花费比较        }        cout<<dp[n]<<endl;    }    return 0;}