ACM学习历程——POJ1260 Pearls（动态规划）

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

这是一道dp题，关键在于想出状态转移方程。
对于n个，一开始最初的计算方式来计算的话，n个是独立的个体，而到最后最优的计算方式，是把某几个设为一类。也是就是相当于把n个插板来分成若干集合，其中的价格取集合中最大的。
这样的话，可以看成这个最终状态是分几步实现的。
从纵向看可以，考虑任意前i个的最优。
从横向来看，任意j，从第j个到第i个都有可能合并成一个集合。.
横向的每次生成对i的一个集合。
纵向形成所有的集合。
状态转移方程为：
dp[i] = min{dp[j] + (sum[i]-sum[j])*p[i]) (j < i);

代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <set>
#include <map>
#include <vector>
#include <queue>
#include <algorithm>
#include <stack>
#include <string>
#define inf 0x3fffffff
#define eps 1e-10

using namespace std;

int a[105], p[105];
int sum[105], dp[105];

int main()
{
    //freopen("test.txt", "r", stdin);
    int T, n;
    scanf("%d", &T);
    for (int times = 0; times < T; ++times)
    {
        scanf("%d", &n);
        for (int i = 1; i <= n; ++i)
        {
            scanf("%d%d", &a[i], &p[i]);
        }
        sum[0] = 0;
        dp[0] = 0;
        for (int i = 1; i <= n; ++i)
        {
            sum[i] = sum[i-1] + a[i];
            dp[i] = dp[i-1] + (a[i]+10)*p[i];
        }
        for (int i = 1; i <= n; ++i)
        {
            for (int j = 0; j < i; ++j)
            {
                dp[i] = min(
                            dp[i],
                            dp[j] + (sum[i]-sum[j]+10)*p[i]);
            }
        }
        printf("%d\n", dp[n]);
    }
    return 0;
}