poj1260 Pearls（Dp）

原题点这里 Time Limit: 1000MS Memory Limit: 10000K

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

题目大意

这道题意思就是对于n种珍珠，各自有各自数量和单价（都是升序排列）

每多选一种珍珠会有额外的手续费=该珍珠单价的10倍

现在要收购所有珍珠数量之和那么多的珍珠，

在求最小花费的同时要保证品质不会下降(即上升或不变)

思路

对于这道题，我们可以总结以下几个特点

（1） 要买的珍珠数量是一定的（废话）

（2） 所买的珍珠的质量只允许提高，不允许下降（即可以用高质量珍珠替代低质量）

（3） 后输入的珍珠价格一定比前面输入的要高

综上，可得珍珠的替代一定是连续的，不能隔代替代
（因为假如用第i+2类去替代第i类珍珠，能使最终的支付价格降低，那么用第i+1类去替代第i类珍珠肯定会使最终的价格更低）

这一点非常关键是以为我们就可以考虑动归方程了

我们先设买到该类珍珠的价格是前面的珍珠价格加现在珍珠价格加手续费

再与之前每一次之前的珍珠价格加现在珍珠加手续费比较取最小值

说人话就是

可能从当前-1的时候开始买这类珍珠，也可能从最开始~当前-2的时候开始买这类珍珠

这样说起来感觉和迷之阶梯 那道题有点像

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int M=1005;int t,n;int s[M],sum[M],p[M],f[M];void init()//数组的清零{    memset(s,0,sizeof(s));  //珍珠数目     memset(sum,0,sizeof(sum));  //前缀和数组     memset(p,0,sizeof(p));  //珍珠单价     memset(f,0,sizeof(f));  //dp数组 } void work(){    scanf("%d",&n);    init();    for(int i=1;i<=n;i++)    {        scanf("%d%d",&s[i],&p[i]);        sum[i]=sum[i-1]+s[i]; //前缀和     }     for(int i=1;i<=n;i++)      {          f[i]=(s[i]+10)*p[i]+f[i-1]; //直接买的价格         for(int j=0;j<i;j++)  f[i]=min(f[i],(sum[i]-sum[j]+10)*p[i]+f[j]);//j的时候就一直买i类珍珠的价格     }      printf("%d\n",f[n]);}int main(){    scanf("%d",&t);    while(t--)work();    return 0;}