bzoj3675: [Apio2014] Sequence Segmentation

Slope optimization. If I can't speak clearly, I'll go to the main idea of the Star God.

I started thinking about an O (kn^3).

is to enumerate K and N, then enumerate the preceding points, and then enumerate the previous points to the current point at which to break the optimal.

and O (kn^2) What the hell?

I find out by drawing (actually someone has told me the problem of distributive law)

This is the last example of a sample break.

(4), (1,3), (4,0), (2,3)

And then the answer is that every number and the other different groups multiply the number of times

The DP equation concludes that: f[i][now]=f[j][now^1]+s[j]* (S[i]-s[j]);

Where now is the scroll array scroll enumeration K of T

Then the slope is optimized.

But there is a dog than the problem, that is s[j1]==s[j2], when the divisor of 0 to be a special sentence

The star-sense God said he found it when he tuned the sample.

But my metaphysics has gone through the examples.

#include <cstdio>#include<iostream>#include<cstring>#include<cstdlib>#include<algorithm>#include<cmath>using namespaceStd;typedefLong LongLL; LL a[110000],s[110000],f[110000][2];intNow ;DoubleYintj) {return Double(f[j][now^1]-s[j]*s[j]);}DoubleXintj) {return Double(S[j]);}DoubleSlopeintJ1,intJ2) {return(Y (J1)-y (J2))/(X (J2)-X (J1));}inthead,tail,list[110000];intMain () {intn,m; scanf ("%d%d",&n,&m); s[0]=0;  for(intI=1; i<=n;i++) scanf ("%lld", &a[i]), s[i]=s[i-1]+A[i]; Memset (F,0,sizeof(f)); now=0;  for(intt=1; t<=m;t++)//Section T cut{ Now^=1; Head=1, tail=1; list[head]=T;  for(inti=t+1; i<=n;i++)        {             while(head!=tail&& (Slope (list[head],list[head+1]) <Double(S[i]) | | S[list[head]]==s[list[head+1]]) head++; intj=List[head]; F[i][now]=f[j][now^1]+s[j]* (s[i]-S[j]);  while(head!=tail&& (s[i]==s[list[tail]]| | Slope (list[tail-1],list[tail]) >slope (list[tail],i)) tail--; list[++tail]=i; }} printf ("%lld\n", F[n][now]); return 0;}

bzoj3675: [Apio2014] Sequence Segmentation