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
不应该排序的，排完序就错了。
dp[i]：存放购买前i个Pearls的最优解。
则有dp[i]=min(dp[j]+(s[i]-s[j]+10)*p[i],dp[i])
#include <stdio.h> #include <iostream> using namespace std; int main() { int t,c; int a[1001]; int p[1001]; int dp[1001]; scanf("%d",&t); while(t--){ int s[1001]={0}; scanf("%d",&c); for(int i=0; i<c; i++){ scanf("%d %d" ,&a[i] ,&p[i]); if(i==0){ s[i]=a[i]; }else{ s[i]=s[i-1]+a[i]; } dp[i]=(s[i]+10)*p[i]; } for(int i=0; i<c; i++){ for(int j=0; j<=i; j++){ dp[i]=min( dp[j]+(s[i]-s[j]+10)*p[i] , dp[i] ); } } printf("%d\n",dp[c-1]); } return 0; }
原文：http://www.cnblogs.com/chenjianxiang/p/3561657.html