斜率dp cdq 分治

f[i] = min { f[j] + sqr(a[i] - a[j]) }

f[i]= min { -2 * a[i] * a[j] + a[j] * a[j] + f[j] } + a[i] * a[i]

由于a[i]不是单调递增的，不能直接斜率dp。

考虑有cdq分治来做，复杂度(nlog2n)

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
 
#define maxn 100008
#define LL long long 

long long f[maxn];
int a[maxn],b[maxn];
int n;
bool flag=1;

void read(int &x){
    char ch;
    for (ch=getchar();ch<'0'||ch>'9';ch=getchar()); x=ch-48;
    for (ch=getchar();ch>='0'&&ch<='9';ch=getchar()) x=x*10+ch-48;
}
 
void init(){
    read(n);
    for (int i=1;i<=n;i++) { read(a[i]); read(b[i]); if (b[i]) flag=0; }
    for (int i=1;i<=n;i++) f[i]=(LL)1<<60;
}
 
void force(){
    for (int i=1;i<=n;i++)
        for (int j=0;j<=i-1;j++)
            if (a[j]>=b[i])
                f[i]=min(f[i],f[j]+(LL)(a[i]-a[j])*(a[i]-a[j]));
    
}
 
 int q[maxn],rk[maxn];
bool cmp(int i,int j){
    return a[i]<a[j] ;
}

long long kx(int i,int j){
    return 2*(a[i]-a[j]);
}

long long ky(int i,int j){
    long long ans=1LL*a[i]*a[i]+f[i]-1LL*a[j]*a[j]-f[j]; 
    return ans;
}

bool cmp1(int i,int j,int k){
    return ky(k,j)*kx(j,i)<=kx(k,j)*ky(j,i);
}

bool cmp2(int i,int j,int k){
    return ky(i,j)>=k*kx(i,j); 
}

 void solve(int l,int r){
    if (l==r) return; 
    int mid=(l+r)>>1;
    solve(l,mid);
    for (int i=l;i<=r;i++) rk[i]=i;
    sort(rk+l,rk+r+1,cmp);
    int h=1,t=0;
    for (int i=l;i<=r;i++)
    {
        if (rk[i]<=mid) {
            while (h<t&&cmp1(q[t-1],q[t],rk[i])) t--;
            q[++t]=rk[i];
        } else {
            while (h<t&&cmp2(q[h],q[h+1],a[rk[i]])) h++;
            f[rk[i]]=min(f[rk[i]],f[q[h]]+1LL*(a[rk[i]]-a[q[h]])*(a[rk[i]]-a[q[h]]));
        }
    }
    solve(mid+1,r);
 }
int main(){
    init();
    if (n<=1000) force(); 
    if (flag) solve(0,n);
    printf("%.4f",sqrt(f[n]));
}