題意： n件物品，給出數(shù)量和價(jià)格，（注意數(shù)量和價(jià)格都是升序給出的這個(gè)是能DP的關(guān)鍵），要買(mǎi)掉所以商品 對(duì)于每類(lèi)物品，所需要的價(jià)格是(a[i]+10)*p[i] ，即要多買(mǎi)10件，也可以把價(jià)格低的物品合并到高價(jià)格的物品那 即 (a[j]+a[i]+10)*p[j];代表把價(jià)格為i的物品合并到j(luò)上。設(shè)dp[i]為買(mǎi)完第i類(lèi)物品的最優(yōu)解 ? ? ?dp[i]=min(dp[i],dp[j]+(a[j+1]+a[j+2]+a[i]+10)*p[i]);
即對(duì)于第i類(lèi)物品來(lái)說(shuō)，它最多合并前面的i-1種物品，而且因?yàn)槲锲窋?shù)目和價(jià)格都是升序的，所以可以證明只要 第j類(lèi)物品能合并，那么再往后就都能合并，因?yàn)榈趈類(lèi)物品合并省出的錢(qián)可以表示為 a[j]*p[i]-(a[j]+10)*p[j] ;如果上式小于0，說(shuō)明可以合并j，而當(dāng)j增大，數(shù)目和價(jià)格都增大，所以只會(huì)比j的時(shí)候更加省錢(qián)，即更能替換。
綜上，有兩種辦法：從前往后推，狀態(tài)轉(zhuǎn)移方程上面以給出。（主函數(shù)）

Time limit? ?1000 ms

Memory limit? ? ?10000 kB

OS? ? ?Linux

Source? ? ?Northwestern Europe 2002

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從后往前推。因?yàn)閷?duì)于最后一種狀態(tài)n，要從前面的n-1種狀態(tài)中倒著連續(xù)合并。（開(kāi)函數(shù)）如下 #include<iostream> #include<stdio.h> #include<string.h> using namespace std; int n,dp[10010]; struct node {int a,b,u,v; } s[10010]; int dfs(int step) {if(step<1)return 0;if(dp[step]!=-1)return dp[step];if(step==1)return dp[step]=s[step].u;dp[step]=s[step].u+dfs(step-1);for(int i=step-1; i>=1; i--)dp[step]=min(dp[step],(s[step].v-s[i-1].v+10)*s[step].b+dfs(i-1));return dp[step]; } int main() {int m;cin>>m;while (m--){cin>>n;s[0].v=0;memset(dp,-1,sizeof(dp));for(int i=1; i<=n; i++){cin>>s[i].a>>s[i].b;s[i].u=(s[i].a+10)*s[i].b;s[i].v=s[i-1].v+s[i].a;}cout<<dfs(n)<<endl;}return 0; }

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