acm pku 1260 Pearls的动态规划算法实现

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Pearls

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

2
2
100 1
100 2
3
1 10
1 11
100 12

Sample Output

330
1344

Source

Northwestern Europe 2002

 

分析:这道题目的大概意思如下:对具有“cnum”个等级的珍珠,根据需求怎样购买才能使得费用最小,低等级的珍珠可以以高等级的珍珠的价格进行购买。分析后,可得出如下两个结论:

1           对任意等级i和等级ji<j)的j-i+1种相邻珍珠,若将珍珠i(等级为i的珍珠)当做珍珠j来购买(Res[i][j]),则ij之间的所有珍珠都应当做珍珠i来处理,否则,必定存在着k(i<k<j)k没有当做j来购买,则可将i当做k来购买,这样更省。

2           购买从任意等级i到等级ji<j)的j-i+1种相邻珍珠时,其最优购买花费(MinRes[i][j])具有如下格式:

MinRes[i][j] = min{MinRes[i][k]+Min[k+1][j](i<=k<j), Res[i][j]}

 

代码实现:

#include <string.h>

#include "iostream"

using namespace std;

 

#define N 100

 

int Res[N][N]; //Res[i][j](i<j) is the total cost of pearls from class i to class j are bought in the jth class price;

int ai[N], pi[N];

int MinRes[N][N]; //MinRes[i][j](i<j) is the minimum cost of pearls from class i to class j, and RES[0][N-1] is what we search for;

 

void GetRes(int cnum)

{

       int i, j, k;

       int snum;

 

       for(i = 0; i < cnum; i++)

              memset(Res[i], 0, sizeof(int)*cnum);

       for(i = 0; i < cnum; i++)

       {

              for(j = i; j < cnum; j++)

              {

                     snum = 0;

                     for(k = i; k <= j; k++)

                            snum += ai[k];

                     Res[i][j] = (snum + 10) * pi[j];

              }

       }

}

 

void GetMinRes(int cnum)

{

       GetRes(cnum);

 

       int i, j, k;

       int imin;

 

       for(i = 0; i < cnum; i++)

              memset(MinRes, 0, sizeof(int)*cnum);

       for(i = cnum-1; i > -1; i--)

       {

              for(j = i; j < cnum; j++)

              {

                     imin = Res[i][j];

                     for(k = i; k < j; k++)

                     {

                            if(imin > MinRes[i][k] + MinRes[k+1][j])

                                   imin = MinRes[i][k] + MinRes[k+1][j];

                     }

                     MinRes[i][j] = imin;

              }

       }

}

 

int main(void)

{

       int n, cnum;

       int i;

 

       cin >> n;

       while(n != 0)

       {

              cin >> cnum;

              for(i = 0; i < cnum; i++)

              {

                     cin >> ai[i] >> pi[i];

              }

              GetMinRes(cnum);

              cout << MinRes[0][cnum-1] << endl;

              n-- ;

       }

       return 0;

}

 

提交结果:

Problem: 1260

 

User: uestcshe

Memory: 320K

 

Time: 16MS

Language: C++

 

Result: Accepted

 

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