UOJ104 "APIO2014" Split the sequence

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Title Link: http://uoj.ac/problem/104

Positive solution: dp+ slope optimization

Problem Solving Report:

　　easy to find, the answer is only related to the division, and the order is irrelevant.

So the naïve equation is easy to get:

Make ${s[n]=\sum_{i=1}^{n}a[i]}$

${f[i][k]=max (f[j][k-1]+s[j]* (s[i]-s[j)), j<i}$

For ${j1,j2}$ and satisfies ${j1<j2}$,${f[j1][k-1]<f[j2][k-1]}$, obviously $j1$ can be deleted, then

${f[j1][k-1]+s[j1]* (S[i]-s[j1]) < f[j2][k-1]+s[j2]* (S[i]-s[j2])}$

After simplification:

${f[j1][k-1]-f[j2][k-1]+s[j2]^2-s[j1]^2 > s[i]* (s[j2]-s[j1])}$

Make ${g[i][k]=f[i][k]-s[i]^2}$

Then ${g[j1][k-1]-g[j2][k-1]>s[i]* (s[j2]-s[j1])}$

By this step, the positive solution is ready. Obviously we can use the slope optimization + monotone queue, the DP optimization to $o (NK) $, do k times, each time just sweep once.

If the first team satisfies the above formula, it is deleted directly. At the end of the team, look at the trend of slope change, if not satisfied then pop off.

　　

It is made by Ljh2000#include <iostream> #include <cstdlib> #include <cstring> #include <cstdio > #include <cmath> #include <algorithm> #include <vector> #include <queue> #include < Complex>using namespace Std;typedef long long ll;const int maxn = 100011;int n,k,dui[maxn],head,tail; LL g[maxn],s[maxn],f[maxn],f[maxn];int ans[maxn],pre[maxn][211];inline int getint () {int w=0,q=0; char C=getchar (); whi Le ((c< ' 0 ' | |    C> ' 9 ') && c!= '-') C=getchar (); if (c== '-') Q=1,c=getchar (); while (c>= ' 0 ' &&c<= ' 9 ') w=w*10+c-' 0 ', C=getchar (); return q?-w:w;} Inline LL calc (int i,int j1,int J2) {return s[i]* (s[j2]-s[j1]);} inline void work () {n=getint (); K=getint (), for (int i=1;i<=n;i++) S[i]=getint (), for (int i=2;i<=n;i++) s[i]+=s[i-1 ]; for (int i=1;i<=n;i++) g[i]=-s[i]*s[i];for (int nowk=2;nowk<=k+1;nowk++) {head=tail=0;dui[++tail]=nowk-1; int i=nowk;i<=n;i++) {while (Head<tail && calc (i,dui[head],dui[head+1])>= (g[dui[head]]-g[dui[head+1])) head++; f[i]=f[dui[head]]+s[dui[head]]* (S[i]-s[dui[head]); Pre[i][nowk-1]=dui[head];while (Head<tail && (G[dui[tail-1]]-g[dui[tail]) * (S[i]-s[dui[tail]]) >= (g[ Dui[tail]]-g[i]) * (S[DUI[TAIL]]-S[DUI[TAIL-1])) tail--;d ui[++tail]=i;} for (int i=nowk;i<=n;i++) f[i]=f[i],g[i]=f[i]-s[i]*s[i];} printf ("%lld\n", F[n]); for (int i=n,j=k;i>0;j--) i=pre[i][j],ans[j]=i;for (int i=1;i<=k;i++) printf ("%d", Ans[i] );}    int main () {work (); return 0;}

　　

　　

UOJ104 "APIO2014" Split the sequence