POJ1260-Pearls(dp)

POJ1260 原题链接http://poj.org/problem?id=1260

Pearls

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 9562 Accepted: 4873

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

22100 1100 231 101 11100 12

Sample Output

3301344

题目大意:给出n种珍珠的数量与价格,要求用最少的钱买到相同数量的质量相同（或更好的）珍珠,规定买任一类的珍珠n个(价格为p)，都要支付(n+10)*p的钱，即额外支付10*p;

拿样例2来做比喻:

1 10   //买第一类1个 单价10元

1 11   //买第二类1个 单价11元

100 12   //买第三类100个 单价 12元

按常规支付为 (1+10)*10 + (1+10)*11 + (100+10)*12 = 1551元（一共买了102个珍珠）

若全部都买第三种所需价格为:(100+1+1+10)*12=1344元 （同样是买了102个且质量更高）

思路:

本题关键点在于：

（1）       要求要买的珍珠的数量是一定的

（2）       所买的珍珠的质量允许提高，但不允许下降（即可以用高质量珍珠替代低质量）

（3）       输入时，后输入的珍珠价格一定比前面输入的要贵

（4）       由（2）（3）知，珍珠的替代必须是连续的，不能跳跃替代（这个不难证明，因为假如用第i+2类去替代第i类珍珠，会使最终的支付价格降低，那么用第i+1类去替代第i类珍珠会使最终的支付价格更加低）

 

根据这4个约束条件，那么购买珍珠的方案为：

在珍珠类型的总区间[1,c]中划分多个子区间，其中在闭区间i1~j1的珍珠全部按第j1类珍珠的价格p1支付，在闭区间i2~j2的珍珠全部按第j2类珍珠的价格p2支付，…在闭区间in~jn的珍珠全部按第jn类珍珠的价格pn支付。 这些区间互不相交。

其余珍珠按其原价支付。

要求找出最优的划分方案，使得最终支付价格最低。

 

令dp[i]表示在已知第i类珍珠时,所需支付的最低价格

则状态方程为：

dp[i]=(a[i]+10)*p[i]+dp[i-1];  //当第i种珍珠出现时，未优化价格的情况

dp[i]=min(dp[i],(sum[i]-sum[j]+10)*p[i]+dp[j]);  //枚举j，价格优化

 

dp[0]=0;  //Dp初始化

代码如下:

#include<stdio.h>#include<string.h>#define min(a,b) a>b?b:aint sum[1100];int num[1100];int price[1100];int dp[1100];int m,n;int i,j,k;int main(){    int T;    scanf("%d",&T);    while(T--)    {        memset(num,0,sizeof(num));//珍珠数目        memset(sum,0,sizeof(sum));//前i中珍珠的所有和，用来求差        memset(price,0,sizeof(price));//珍珠的价格        memset(dp,0,sizeof(dp));//表示前i中珍珠的最优解，即支付的最低价格        scanf("%d",&n);        sum[0]=0;        for(i=1;i<=n;i++)        {            scanf("%d%d",&num[i],&price[i]);            sum[i]=sum[i-1]+num[i];        }        dp[0]=0;//dp的初始化        for(i=1;i<=n;i++)        {            dp[i]=(num[i]+10)*price[i]+dp[i-1];//当第ii种珍珠出现 未求最优解的情况             for(j=0;j<i;j++)//枚举第i种前的所有珍珠 被i所替代的情况寻找最优解             {                 dp[i]=min(dp[i],(sum[i]-sum[j]+10)*price[i]+dp[j]);//(sum[i]-sum[j]+10)*p[i]+dp[j]表示j+1到第i种被i取代的价格             }        }        printf("%d\n",dp[n]);    }    return 0;}