BZOJ[4709][Jsoi2011]柠檬 斜率优化

传送门ber~
容易发现选的每一段左右一定是同色…
那么有$f_i=max\{f_{j-1}+(s_i-s_j+1)^2*a_i\}$
其中$i$,$j$同色
若存在$k>t$且答案更优，即

$$f_{k-1}+(s_i-s_k+1)^2*a_i>f_{t-1}+(s_i-s_t+1)^2*a_i$$
$$f_{k-1}-f_{t-1}>a_i*(s_i^2+s_t^2+1-2s_is_t-2s_t+2s_i-(s_i^2+s_k^2+1-2s_is_k-2s_k+2s_i))$$
$$f_{k-1}-f_{t-1}>a_i*(s_t^2-2s_is_t-2s_t-s_k^2+2s_is_k+2s_k)$$
$$f_{k-1}-f_{t-1}>ps_t^2-2ps_is_t-2ps_t-ps_k^2+2ps_is_k+2ps_k$$
$$f_{k-1}-f_{t-1}+ps_k^2-ps_t^2-2ps_k+2ps_t>s_i(2ps_k-2ps_t)$$
$$\frac {f_{k-1}-f_{t-1}+ps_k^2-ps_t^2-2ps_k+2ps_t}{2ps_k-2ps_t}>s_i$$
$y=f_{k-1}+ps_k^2-2ps_k$
$x=2ps_k$
$f_{k-1}+ps_i^2+ps_k^2+p-2ps_is_k-2ps_k+2ps_i$
$=y-xs_i+ps_i^2+p+2ps_i$
斜率优化
单调栈
代码如下·：

#include<algorithm>
#include<cstring>
#include<ctype.h>
#include<cstdio>
#include<vector>
#define int long long
#define INF 2147483647
#define N 100020
using namespace std;
inline int read(){
    int x=0,f=1;char c;
    do c=getchar(),f=c=='-'?-1:f; while(!isdigit(c));
    do x=(x<<3)+(x<<1)+c-'0',c=getchar(); while(isdigit(c));
    return x*f;
}
typedef long long LL;
struct Data{
    LL x,y;
    Data(){}
    Data(LL x,LL y):x(x),y(y){}
}tmp;
vector<Data>s[N];
int n,x,y,p,ans;
int cnt[N],pos[N],a[N],pre[N],top[N],f[N];
inline double Slope(Data a,Data b){
    if(a.x==b.x) return a.y>b.y?-INF:INF;
    return 1.0*(b.y-a.y)/(b.x-a.x);
}
 main(){
    n=read();
    for(int i=1;i<=n;i++){
        a[i]=read();
        pre[i]=pos[a[i]];
        pos[a[i]]=i;
        cnt[i]=cnt[pre[i]]+1;
    }
    memset(pos,0,sizeof pos);
    for(int i=1;i<=n;++i)
        if(!pos[a[i]])
            s[a[i]].push_back(Data(0,0)),pos[a[i]]=1;
    for(int i=1;i<=n;i++){
        p=a[i];
        tmp=Data(2*p*cnt[i],f[i-1]+p*cnt[i]*cnt[i]-2*p*cnt[i]);
        while(top[p]>1 && Slope(s[p][top[p]-1],s[p][top[p]])<=Slope(s[p][top[p]],tmp)) top[p]--,s[p].pop_back();
        top[p]++;s[p].push_back(tmp);
        while(top[p]>1 && Slope(s[p][top[p]-1],s[p][top[p]])<=cnt[i]) top[p]--,s[p].pop_back();
        x=s[p][top[p]].x;y=s[p][top[p]].y;
        f[i]=y-x*cnt[i]+p*cnt[i]*cnt[i]+p+2*p*cnt[i];
        ans=max(ans,f[i]);
    }
    printf("%lld",ans);
    return 0;
}