BZOJ3675 & Rokua 3648 & uoj104:[apio2014] Sequence Segmentation--the puzzle

https://www.lydsy.com/JudgeOnline/problem.php?id=3675

https://www.luogu.org/problemnew/show/P3648

http://uoj.ac/problem/104

PS: The problem surface and the solution for Ullo and uoj version, Bzoj please consciously the "output procedure" deleted.

Little H has recently hooked up to a game of delimited sequences. In this game, small h needs to divide a series of non-negative integers of length n into k+1 non-empty subsequence sequences. In order to get the k+1 subsequence, the small H needs to repeat the following steps of K: 1. Small h first selects a sequence longer than 1 (at the beginning of the small h only a sequence of length n-that is, the entire sequence obtained at the beginning); 2. Select a location, And through this position the sequence is divided into two consecutive non-empty new sequences. After each of these steps, small h will get a certain score. This fraction is the product of the elements and in two new sequences. Small h wants to choose the best way to divide, so that after K-wheel, the total score of small h is the largest.

Reference: Luo gu (although not counted as reference emmm just slope optimization write kneeling to debug with).

It is not difficult to testify that the answer is fixed so long as the split point is fixed, so O (kn^2) is not difficult to think.

Make s[i] represent the prefix of the first I and then we have:

F[k][i]=max (f[k][i],f[k-1][j]+ (S[i]-s[j]) *s[j])

This equation can obviously be optimized for slope, maintaining a monotone non-additive sequence that makes k<j<i.

Ignoring the previous dimension of F, when f[k]+ (S[i]-s[k]) *s[k]<=f[j]+ (S[i]-s[j]) *s[j], the K is ejected from the first team.

So g[k,j]= (F[K]-F[J]+SQR (S[j])-SQR (S[k])/(S[j]-s[k]) <=s[i] will pop up K.

Also consider G[k,j]>=g[j,i] when the J ejected, give proof.

Obviously when G[j,i]<=s[i] J is not excellent.

When G[k,j]>=g[j,i]>s[i] K better than J still eject J.

I also naive want to maintain monotonous non-reduction sequence results of various wonderful emmm error ...

#include <cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespaceStd;typedefLong LongLl;typedefDoubleDL;Const intn=1e5+5;Const intk=205; inlineintRead () {intx=0, w=0;CharCh=0;  while(!isdigit (CH)) {w|=ch=='-'; ch=GetChar ();}  while(IsDigit (CH)) X= (x<<3) + (x<<1) + (ch^ -), ch=GetChar (); returnw?-x:x;} ll f[2][n];intn,k,nxt[n][k],q[n],s[n],l,r,now=1, pre=0; inline ll Sqr (ll K) {returnk*K;} Inline DL Suan (intJintk) {    if(S[j]==s[k])return-1e18; intI=Pre; return(F[I][K]-F[I][J]+SQR (S[j])-SQR (S[k])/(DL) (s[j]-s[k]);}intMain () {n=read (), k=read ();  for(intI=1; i<=n;i++) s[i]=s[i-1]+read ();  for(intj=1; j<=k;j++) { now^=1, pre^=1; l=r=0;  for(intI=1; i<=n;i++){             while(L<r&&suan (q[l],q[l+1]) <= (DL) s[i]) l++; intt=Q[l]; F[now][i]=f[pre][t]+ (LL) (S[i]-s[t]) *s[t];nxt[i][j]=T;  while(L<r&&suan (q[r-1],q[r]) >=suan (q[r],i)) r--; q[++r]=i; }} printf ("%lld\n", F[now][n]); inttmp=Nxt[n][k];  while(k) {printf ("%d", TMP); TMP=nxt[tmp][--K]; }    return 0;}

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BZOJ3675 & Rokua 3648 & uoj104:[apio2014] Sequence Segmentation--the puzzle