Necklace

frog has \(n\) gems arranged in a cycle, whose beautifulness are \(a_1, a_2, \dots, a_n\). She would like to remove some gems to make them into a beautiful necklace without changing their relative order.

Note that a beautiful necklace can be divided into \(3\) consecutive parts \(X, y, Z\), where

1. \(X\) consists of gems with non-decreasing beautifulness,
2. \(y\) is the only perfect gem. (A perfect gem is a gem whose beautifulness equals to \(10000\))
3. \(Z\) consists of gems with non-increasing beautifulness.

Find out the maximum total beautifulness of the remaining gems.

Input

The input consists of multiple tests. For each test:

The first line contains \(1\) integer \(n\) (\(1 \leq n \leq 10^5\)). The second line contains \(n\) integers \(a_1, a_2, \dots, a_n\). (\(0 \leq a_i \leq 10^4\), \(1 \leq \textrm{number of perfect gems} \leq 10\)).

Output

For each test, write \(1\) integer which denotes the maximum total remaining beautifulness

Sample Input

    6
    10000 3 2 4 2 3
    2
    10000 10000

Sample Output

    10010
    10000

#include <bits/stdc++.h>
#define  met(a,b) memset(a,b,sizeof a)
using namespace std;
typedef long long ll;
const int N = 1e5+50;
int C[N];
int dp1[N],dp2[N],b[N],a[2*N];
int n,m,k;
int low_bit(int x)
{
    return x&(-x);
}
void updata(int pos,int val)
{
    for(int i=pos;i<=10000;i+=low_bit(i))
        C[i]=max(C[i],val);
}
int get_max(int pos)
{
    int ans=0;
    for(int i=pos;i>0;i-=low_bit(i))
        ans=max(ans,C[i]);
    return ans;
}

int solve(int id)
{
    met(C,0);
    int cnt=0;
    for(int i=id+1;i<id+n;i++)
    {
        if(a[i]!=10000)
            b[cnt++]=a[i];
    }
    dp1[0]=b[0];
    updata(10000-b[0],b[0]);
    for(int i=1;i<cnt;i++)
    {
        dp1[i]=get_max(10000-b[i])+b[i];
        updata(10000-b[i],dp1[i]);
    }
    met(C,0);
    dp2[cnt-1]=b[cnt-1];
    updata(10000-b[cnt-1],b[cnt-1]);
    for(int i=cnt-2;i>=0;i--)
    {
        dp2[i]=get_max(10000-b[i])+b[i];
        updata(10000-b[i],dp2[i]);
    }
    int ans=0;
    for(int i=1;i<cnt;i++)
        dp1[i]=max(dp1[i],dp1[i-1]);
    for(int i=cnt-2;i>0;i--)
        dp2[i]=max(dp2[i],dp2[i+1]);
    dp2[cnt]=0;
    for(int i=0;i<cnt;i++)
        ans=max(ans,dp1[i]+dp2[i+1]);
    ans=max(ans,dp2[0]);
    return ans+10000;
}
int main()
{
    while(~scanf("%d",&n))
    {
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
            a[i+n]=a[i];
        }
        int ans=0;
        for(int i=0;i<n;i++)
        {
            if(a[i]==10000)
            {
                ans=max(ans,solve(i));
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}

SCU - 4441 Necklace（樹狀數組求最長上升子數列）