HDU 4960(Another OCD Patient-区间dp)

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∑ki=1a[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
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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;
}