题解：P5504 [JSOI2011] 柠檬

题目：

下面给个经典的 DP 式子不多说了：

\[f_i=f_j+s_iqz^2(s_i,i)+s_iqz^2(s_{j+1},j)-2s_iqz(s_i,i)qz(s_{j+1},j),s_{j+1}=s_i \]

单调栈太阴了！下面有个 hack：

11
2 10 2 2 10 2 2 10 2 2 2
ans：128

众所周知 \(x,k\) 都单调的题我们可以用单调队列维护。
但是这题是上凸包且 \(k\) 递增，手画一下，我们会发现我们实际上决策点是从右端点开始缩。总的看下来是个单调栈。

#include<bits/stdc++.h>
#define int long long
using namespace std;
const int QAQ=1e5+7,ovo=1e4+7;
int n,s[QAQ];
vector<int> qz[ovo];
int qzh(int s,int i) {return upper_bound(qz[s].begin(),qz[s].end(),i)-qz[s].begin();}
struct dian {int x,y;} ;
vector<dian> zhan[ovo];
const double inf=1e18;
double xie(dian a,dian b)
{if(a.x==b.x) return (b.y>=a.y)?inf:-inf;return (1.0*b.y-1.0*a.y)/(1.0*b.x-1.0*a.x);
}
int l[ovo],r[QAQ],f[QAQ],ans;
signed main()
{cin>>n;for(int i=1;i<=n;i++) cin>>s[i],qz[s[i]].push_back(i);for(int i=1;i<=10000;i++) zhan[i].push_back({0,0});for(int i=1;i<=n;i++){ int ccx=qzh(s[i],i);int k=2*s[i]*ccx;while(l[s[i]]<r[s[i]]&&xie(zhan[s[i]][r[s[i]]-1],zhan[s[i]][r[s[i]]])<=k)r[s[i]]--;f[i]=s[i]*ccx*ccx+zhan[s[i]][r[s[i]]].y-k*zhan[s[i]][r[s[i]]].x;int swl=qzh(s[i+1],i);dian nw={swl,f[i]+s[i+1]*swl*swl};while(l[s[i+1]]<r[s[i+1]]&&xie(zhan[s[i+1]][r[s[i+1]]-1],zhan[s[i+1]][r[s[i+1]]])<=xie(zhan[s[i+1]][r[s[i+1]]-1],nw))r[s[i+1]]--;++r[s[i+1]];if(zhan[s[i+1]].size()<=r[s[i+1]]) zhan[s[i+1]].push_back(nw);else zhan[s[i+1]][r[s[i+1]]]=nw;ans=max(ans,f[i]);}cout<<ans;return 0;
}