Another OCD Patient
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1320 Accepted Submission(s): 459

Problem Description
已知一个数列，现在把它分成k段，要求每段的和组成的新数列回文，且合并代价最小。
合并的代价为$\sum_{i=1}^{k}a[len](len为该段的长度)$

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

HintIn 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.

Author
SYSU

Source
2014 Multi-University Training Contest 9

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#include<bits/stdc++.h>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define pb push_back
#define mp make_pair 
#define MAXN (5000+10)
typedef __int64 ll;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int n,a[MAXN],v[MAXN];
ll s[MAXN],s2[MAXN];
int l[MAXN],r[MAXN],tot=0;
ll cost[MAXN];
int main()
{
//  freopen("A.in","r",stdin);
//  freopen(".out","w",stdout);while(scanf("%d",&n)&&n) {For(i,n) scanf("%d",&v[i]);For(i,n) scanf("%d",&a[i]); a[0]=0;s[0]=s2[n+1]=0;For(i,n) s[i]=s[i-1]+(ll)v[i];ForD(i,n) s2[i]=s2[i+1]+(ll)v[i];tot=0; int fro=1,tail=n;while(fro<tail) {if (s[fro]==s2[tail]) {l[++tot]=fro;r[tot]=tail;fro++,tail--;} else if (s[fro]<s2[tail]) ++fro;else --tail;}MEMI(cost)l[0]=0,r[0]=n+1; cost[0]=0;ll ans=a[n];For(i,tot) {Rep(j,i) {cost[i]=min(cost[i],cost[j]+a[l[i]-l[j]]+a[r[j]-r[i]]);}           ans=min(ans,cost[i]+a[r[i]-1-l[i]]) ;       } cout<<ans<<endl;} return 0;
}