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
4000
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i到j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

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <queue>
#include <cstdio>
#include <map>
using namespace std;
#define LL long long int
LL pre[5005];
int cost[5005];
int dp[5005][5005];
int n;
int dfs(int l,int r)
{
if(l>r) return 0;
if(dp[l][r]!=-1) return dp[l][r];
dp[l][r]=cost[r-l+1];//假设将这个区间全部合并

int ll=l,rr=r;
while(ll<rr)
{
LL sum1=pre[ll]-pre[l-1];
LL sum2=pre[r]-pre[rr-1];
if(sum1<sum2)
{
ll++;
continue;
}
if(sum1>sum2)
{
rr--;
continue;
}
if(sum1==sum2)
{
dp[l][r]=min(dp[l][r],cost[ll-l+1]+cost[r-rr+1]+dfs(ll+1,rr-1));
ll++;
}
}
return dp[l][r];
}
int main()
{
while(scanf("%d",&n)!=EOF&&n)
{
pre[0]=0;
for(int i=1;i<=n;i++)
{
scanf("%lld",&pre[i]);
pre[i]+=pre[i-1];
for(int j=1;j<=n;j++)
dp[i][j]=-1;//初始化dp数组
}
for(int i=1;i<=n;i++)
scanf("%d",&cost[i]);

printf("%d\n",dfs(1,n));
}
return 0;
}