HDU-1300(基础方程DP-遍历之前所有状态)

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

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

 

Sample Output

330 1344

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思路：

之前想了一种思路被验证是想偏了，后来看了正确的答案后发现和HDU的2059龟兔赛跑是同一类型的题目

这俩题耽误了能有两天的时间，不过也算让我对DP的认识又上了一个台阶，因为他们都反映了DP的本质：一个状态的最优值是通过枚举所有能够到达这个状态的状态的值所得到的，知道了这点后，再就具体问题具体分析就OK

下面有两段代码，前一段是是按照之前的思路做的，自己和AC的代码比对了很多组数据（故意挑了一些比较刁钻的数据）结果都是相同的，但不知道为什么还是WA，先放在这儿等以后有能力了再来解决，后面的那段代码就是AC的了。

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 解法一：（WA但数据测试正确）

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

int T;
int number;
struct P{
    __int64 v;
    __int64 c;
}pearl[];
int exits[];
__int64 sum;

int main()
{
    scanf("%d",&T);
    while(T--)
    {
        sum = ;
        scanf("%d",&number);
        for(int i = ;i <= number;i++){
            scanf("%I64d%I64d",&pearl[i].c,&pearl[i].v);
            exits[i] = ;
        }
        for(int i = ;i <= number;i++)
            if(pearl[i-].c*pearl[i].v < (pearl[i-].c+)*pearl[i-].v){
                pearl[i].c += pearl[i-].c;
                exits[i-] = ;
            }
        for(int i = ;i <= number;i++)
            if(exits[i]) sum += (pearl[i].c+)*pearl[i].v;
        printf("%I64d\n",sum);
    }
    return ;
}

解法二（AC）：

#include <iostream>
#include <cstdio>
#include <cstring>
#define INF 0x7fffffff
using namespace std;

int min(int a,int b)
{
    return a<b?a:b;
}

int main()
{
    int T;
    int sum[];
    int val[];
    int dp[];
    int num[];
    scanf("%d",&T);
    while(T--)
    {
        int n;
        scanf("%d",&n);
        sum[] = ;
        for(int i = ;i <= n;i++) {
            scanf("%d%d",&num[i],&val[i]);
            sum[i] = sum[i-]+num[i];
        }
        for(int i = ;i <= n;i++)
            dp[i] = INF;
        sum[] = ;
        dp[] = ;
        for(int i = ;i <= n;i++)
            for(int j = ;j < i;j++)
                dp[i] = min(dp[i],dp[j]+(sum[i]-sum[j]+)*val[i]);
        printf("%d\n",dp[n]);
    }
    return ;
}