Another OCD Patient HDU - 4960 (dp+前缀和)
Xiaoji is an OCD (obsessive-compulsive disorder) patient. This morning, his children played with plasticene. They broke the plasticene into N pieces, and put them in a line. Each piece has a volume Vi. Since Xiaoji is an OCD patient, he can’t stand with the disorder of the volume of the N pieces of plasticene. Now he wants to merge some successive pieces so that the volume in line is symmetrical! For example, (10, 20, 20, 10), (4,1,4) and (2) are symmetrical but (3,1,2), (3, 1, 1) and (1, 2, 1, 2) are not.
However, because Xiaoji’s OCD is more and more serious, now he has a strange opinion that merging i successive pieces into one will cost ai. And he wants to achieve his goal with minimum cost. Can you help him?
By the way, if one piece is merged by Xiaoji, he would not use it to merge again. Don’t ask why. You should know Xiaoji has an OCD.
Input
The input contains multiple test cases.
The first line of each case is an integer N (0 < N <= 5000), indicating the number of pieces in a line. The second line contains N integers Vi, volume of each piece (0 < Vi <=10^9). The third line contains N integers ai (0 < ai <=10000), and a1 is always 0.
The input is terminated by N = 0.
Output
Output one line containing the minimum cost of all operations Xiaoji needs.
Sample Input
5
6 2 8 7 1
0 5 2 10 20
0
Sample Output
10
Hint
In the sample, there is two ways to achieve Xiaoji’s goal.
[6 2 8 7 1] -> [8 8 7 1] -> [8 8 8] will cost 5 + 5 = 10.
[6 2 8 7 1] -> [24] will cost 20.
大致题意：就是给你n个数的一个序列，然后你可以将连续的一串数相加合并一个数，告诉你合并1个到n个数的花费，问使得这个序列变成一个回文串所需花费的最少代价。
思路：假设dp[l][r]表示使得l到r这个区间内的序列变成回文串的最小花费，那么状态转移方程为
dp[l][r]=min(dp[l][r],dp[i][j]+cost[i-l+1]+cost[r-j+1]) （ l < i < j < r ,假设l到i区间的数总和等于j到r区间数的总和，那么分别将他们合并，左右两边的数就相等了），这里我们用前缀和来快速求出某个区间的和。
代码如下
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