HDOJ 1300

Pearls

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1173    Accepted Submission(s): 519

Problem 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

 

Source

Northwestern Europe 2002

 

Recommend

Eddy

 

挺简单的一个DP，自己想半天都没想出，看了别人得思路后恍然大悟啊

1.假设共有t个级别的珍珠。解可以被描述成(1,2...i),(i+1,i+2...k)...(s,s+1...t)形式，每个括号中的珍珠按该括号中最后一个级别的珍珠的价格购买。这里要注意的是，假设按等级由低到高的顺序，有i，j，k三个级别的珍珠，答案中不可能出现i按k的价格购买，j按j自己的价格购买的情况。因为如果j按自己的价格购买，那么以j的价格来买i肯定比以k的价格买便宜。所以按高等级的珍珠k的价格购买低等级的珍珠时，低等级的珍珠是连续的，且挨着k。

2.sum是一位数组，sum[i]存放1~i级别的珍珠的需求量之和。ans是一位数组，ans[i]存放购买前i个级别的珍珠的最小开销。ans[i]=min{ans[i],(sum[i]-sum[j]+10)*price[i]+ans[j]}，1<=j<i。

#include <cstdio>#include <cmath>#include <algorithm>#include <iostream>#include <cstring>#include <map>#include <string>#include <stack>#include <cctype>#include <vector>#include <queue>#include <set>#include <iomanip>using namespace std;//#define Online_Judge#define outstars cout << "***********************" << endl;#define clr(a,b) memset(a,b,sizeof(a))#define lson l , mid  , rt << 1#define rson mid + 1 , r , rt << 1|1#define FOR(i , x , n) for(int i = (x) ; i < (n) ; i++)#define FORR(i , x , n) for(int i = (x) ; i <= (n) ; i++)#define REP(i , x , n) for(int i = (x) ; i > (n) ; i--)#define REPP(i ,x , n) for(int i = (x) ; i >= (n) ; i--)#define mk make_pairconst int inf = 1 << 30;const int MAXN = 100 + 50;const int maxw = 30000 + 20;const int MAXNNODE = 1000 +10;const long long LLMAX = 0x7fffffffffffffffLL;const long long LLMIN = 0x8000000000000000LL;const int INF = 0x7fffffff;const int IMIN = 0x80000000;#define eps 1e-8#define mod 20071027typedef long long LL;const double PI = acos(-1.0);typedef double D;typedef pair<int , int> pii;const D ee = 2.718281828459;int num[MAXN] , cost[MAXN] , sum[MAXN] , ans[MAXN];int main(){    //ios::sync_with_stdio(false);#ifdef Online_Judge    freopen("in.txt","r",stdin);    freopen("out.txt","w",stdout);#endif // Online_Judge    int n , T;    cin >> T;    while(T--)    {        scanf("%d" , &n);        clr(ans , 0);        sum[0] = 0;        FORR(i , 1 , n)        {            scanf("%d%d" , &num[i] , &cost[i]);            sum[i] = sum[i - 1] + num[i];            ans[i] = (sum[i] + 10) * cost[i];        }        FORR(i , 1,  n)FOR(j , 1 , i)        {            ans[i] = min(ans[i] , (sum[i] - sum[j] + 10) * cost[i] + ans[j]);        }        printf("%d\n" , ans[n]);    }    return 0;}