BZOJ3675: [Apio2014] Sequence Segmentation

3675: [Apio2014] sequence split time limit:40 Sec Memory limit:128 MB
submit:218 solved:82
[Submit] [Status] Description

Little H has recently been fascinated by a split-sequence game. In this game, small H takes a long
A non-negative integer sequence of degrees N is divided into k+l non-empty subsequence sequences. In order to get the K+l subsequence,
The small H will repeat the following steps seven times:
1. Small h first selects a sequence with a length of more than 1 (at the beginning of the small h only a length of n
Sequence one by one is the whole sequence obtained at the beginning);
2. Select a location and divide the sequence into contiguous two non-empty new
Sequence.
After each of these steps, small h will get a certain score. This is a score of two new sequences
The product of the elements in the column. Small h wants to choose an optimal partitioning scheme, so that after K-wheel (TH),
The total score for small h is the largest.

Input

The first line of the input file contains two integers n and Guineas (K+1≤n).
The second line contains n non-negative integers a1,n2 .... , an (0≤ai≤10^4), which represents the beginning of a small h
to the sequence.

Output

A row contains an integer that is the maximum score a small h can get.

Sample Input7 3
4 1 3 4 0 2 3
Sample Output108
HINT

"Sample description"

In the example, small H can get 108 points with the following 3-wheel operation:

1. -Start small H has a sequence (4,1,3,4,0,2,3). Small H selects the position after the 1th number

Divide the sequence into two parts and get the 4x (1+3+4+0+2+3) =52 points.

2. At the beginning of this round, small H has two sequences: (4), (1,3,4,0,2,3). Small h selection in 3rd number

The position after the word divides the second sequence into two parts and gets (1+3) x (4+0+2+

3) =36 points.

3. At the beginning of this round, small H has three sequences: (4), (1,3), (4,0,2,3). Small h selection in 5th

The position after the number divides the third sequence into two parts and gets (4+0) x (2+3) =

20 points.

After the three-wheeled operation, the small H will get four sub-sequences: (4), (1,3), (4,0), (2,3) and a total of 52+36+20=108 points.

"Data size and scoring"

: Data meets 2≤n≤100000,1≤k≤min (N-1,200).

Source

Exercises

This is a clever conversion: The final score is only related to the last number of sequences, and the order of the points is irrelevant. (Thought for a long time to come out t_t)

Then we can DP, F[i]=min (f[j]+ (S[i]-s[j]) * (N-s[i]+s[j]) slightly deformed, this can be optimized with slope, so the problem is solved.

Concrete deformation can be http://trinklee.blog.163.com/blog/static/238158060201462310236608/

Because the write slope does not have the special enigmatic grading mother ==0 WA 2 times ...

Code:

1#include <cstdio>2  3#include <cstdlib>4  5#include <cmath>6  7#include <cstring>8  9#include <algorithm>Ten   One#include <iostream> A   -#include <vector> -   the#include <map> -   -#include <Set> -   +#include <queue> -   +#include <string> A   at #defineINF 1000000000 -   - #defineMAXN 100000+1000 -   - #defineMAXM 201 -   in #defineEPS 1e-10 -   to #definell Long Long +   - #definePA pair<int,int> the   * #defineFor0 (i,n) for (int i=0;i<= (n); i++) $  Panax Notoginseng #defineFor1 (i,n) for (int i=1;i<= (n); i++) -   the #defineFor2 (i,x,y) for (int i= (x); i<= (y); i++) +   A #defineFor3 (i,x,y) for (int i= (x); i>= (y); i--) the   + #defineMoD 1000000007 -   $ using namespacestd; $   -InlineintRead () -   the { -  Wuyi     intx=0, f=1;CharCh=GetChar (); the   -      while(ch<'0'|| Ch>'9'){if(ch=='-') f=-1; ch=GetChar ();} Wu   -      while(ch>='0'&&ch<='9') {x=Ten*x+ch-'0'; ch=GetChar ();} About   $     returnx*F; -   - } - intn,m,t=0, Q[MAXN]; All f[2][MAXN],S[MAXN]; +InlineDoubleKintIintj) the { -     if(s[i]-s[j]<eps)returninf; $     Else return(Double) (f[1-t][i]-s[i]*s[i]-f[1-T][J]+S[J]*S[J])/(Double) (s[i]-s[j]); the } the   the intMain () the   - { inFreopen ("Input.txt","R", stdin); theFreopen ("output.txt","W", stdout); theN=read (); m=read (); Abouts[0]=0; theFor1 (i,n) s[i]=s[i-1]+read (); theFor1 (i,n) f[0][i]=s[i]* (s[n]-s[i]); the For1 (j,m) +     { -t=1-T; the         intL=1, r=0;Bayi For1 (i,n) the         { the              while(L<r&&k (q[l+1],Q[L]) >s[n]-2*s[i]) l++; -f[t][i]=f[1-t][q[l]]+ (S[i]-s[q[l]) * (s[n]-s[i]+S[q[l]]); -             if(i<j+1) f[t][i]=0; the             if(i==j+1) f[t][i]=f[1-t][i-1]+ (s[i]-s[i-1]) * (s[n]-s[i]+s[i-1]); the              while(L<r&&k (I,q[r]) >k (q[r],q[r-1])) r--; theq[++r]=i; the         } -     } theprintf"%lld\n",f[t][n]>>1); the   the     return 0;94   the}

View Code

Why are thieves slow? Should it be a question of real numbers?

BZOJ3675: [Apio2014] Sequence Segmentation