洛谷T51924 忘情

二分上界有多大开多大 二分上界有多大开多大 二分上界有多大开多大 重要的事情说三遍

又被bright神仙带着做题了

先无脑上wqs二分

我们可以把这个柿子画一下，区间的花费就变成(（sigema(l~r)i s[i]）+1)^2了

那么这个东西经过我艰苦的画柿子证明是满足四边形不等式的

然后就和贞鱼那题一样搞了？然后我就被卡常了qwq囧

其实是自己思维僵化得厉害

上个斜率优化不好吗2333333

 

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long LL;

int n;LL s[110000],f[110000],g[110000];
int h,t,q[110000];
LL Y(int j){return f[j]+s[j]*s[j]-2*s[j];}
LL X(int j){return s[j];}
void check(LL C)
{
    h=1,t=0;q[++t]=0;f[0]=g[0]=0;
    for(int i=1;i<=n;i++)
    {
        while(h<t&&(Y(q[h+1])-Y(q[h]))<=(X(q[h+1])-X(q[h]))*2*s[i])h++;
        f[i]=f[q[h]]+(s[i]-s[q[h]]+1)*(s[i]-s[q[h]]+1)+C;
        g[i]=g[q[h]]+1;
        while(h<t&& (Y(q[t])-Y(q[t-1]))*(X(i)-X(q[t])) >= (Y(i)-Y(q[t]))*(X(q[t])-X(q[t-1])) )t--;
        q[++t]=i;
    }
}
int main()
{
    int K;
    scanf("%d%d",&n,&K);s[0]=0;
    for(int i=1;i<=n;i++)
        scanf("%lld",&s[i]), s[i]+=s[i-1];
        
    LL l=0,r=1e18,ans;
    while(l<=r)
    {
        LL mid=(l+r)/2;
        check(mid);
        if(g[n]>=K)
        {
            ans=f[n]-K*mid;
            l=mid+1;
        }
        else r=mid-1;
    }
    printf("%lld\n",ans);
    return 0;
}

 

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long LL;

int n;
LL s[110000],f[110000],g[110000];
LL val(int l,int r){return (s[r]-s[l]+1)*(s[r]-s[l]+1);} 
LL cost(int j,int i){return f[j]+val(j,i);}//由j这个决策点更新i的花费 
struct node
{
    int l,r,id;
    node(){}
    node(int L,int R,int ID){l=L;r=R;id=ID;}
}q[110000];
void check(LL C)
{
    int h=1,t=0;q[++t]=node(0,n,0);
    for(int i=1;i<=n;i++)
    {
        if(q[h].r<i)h++;
        q[h].l=i;
        f[i]=cost(q[h].id,i)+C;
        g[i]=g[q[h].id]+1;
        
        if(h>t||cost(i,n)<=cost(q[h].id,n))
        {
            while(h<=t&&cost(i,q[t].l)<=cost(q[t].id,q[t].l))t--;
            if(h>t)q[++t]=node(i,n,i);
            else
            {
                int l=q[t].l,r=q[t].r,ans;
                while(l<=r)
                {
                    int mid=(l+r)/2;
                    if(cost(i,mid)>cost(q[t].id,mid))
                    {
                        ans=mid;
                        l=mid+1;
                    }
                    else r=mid-1;
                }
                q[t].r=ans;
                q[++t]=node(ans+1,n,i);
            }
        }
    }
}
int main()
{
    int K;
    scanf("%d%d",&n,&K);s[0]=0;
    for(int i=1;i<=n;i++)
        scanf("%lld",&s[i]), s[i]+=s[i-1];
        
    LL l=0,r=1e18,ans;
    while(l<=r)
    {
        LL mid=(l+r)/2;
        check(mid);
        if(g[n]>=K)
        {
            ans=f[n]-K*mid;
            l=mid+1;
        }
        else r=mid-1;
    }
    printf("%lld\n",ans);
    return 0;
}