题意

题目链接

Sol

\[f[i], f[j] + (h[i] - h[j])^2 + (w[i - 1] - w[j]))\]

然后直接套路斜率优化，发现\(k, x\)都不单调

写个cdq就过了

辣鸡noi.ac居然出裸题&&原题

#include<bits/stdc++.h> 
#define Pair pair<double, double>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
#define int long long 
#define LL long long 
#define db  double
#define Fin(x) {freopen(#x".in","r",stdin);}
#define Fout(x) {freopen(#x".out","w",stdout);}
using namespace std;
const int MAXN = 1e6 + 10, mod = 1e9 + 7, INF = 1e18 + 10;
const double eps = 1e-9;
template <typename A, typename B> inline bool chmin(A &a, B b){if(a > b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline bool chmax(A &a, B b){if(a < b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline LL add(A x, B y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;}
template <typename A, typename B> inline void add2(A &x, B y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);}
template <typename A, typename B> inline LL mul(A x, B y) {return 1ll * x * y % mod;}
template <typename A, typename B> inline void mul2(A &x, B y) {x = (1ll * x * y % mod + mod) % mod;}
template <typename A> inline void debug(A a){cout << a << '\n';}
template <typename A> inline LL sqr(A x){return 1ll * x * x;}
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int N;
struct Sta {
    int id;
    db h, w, x, y, f, ad;
    void Get() {
        x = 2 * h;
        y = f + h * h - w;
    }
}a[MAXN], st[MAXN];
vector<Pair> v;
int comp(const Sta &a, const Sta &b) {
    return a.id < b.id;
}
double GetK(Pair a, Pair b) {
    if((b.fi - a.fi) < eps) return INF;
    return (b.se - a.se) / (b.fi - a.fi);
}

int sid[MAXN], cur;

void GetConvexHull(int l, int r) {
    while(cur) sid[cur--] = 0;
    v.clear();
    for(int i = l; i <= r; i++) {
        double x = a[i].x, y = a[i].y;
        while((v.size() > 1 && ((GetK(v[v.size() - 1], MP(x, y)) < GetK(v[v.size() - 2], v[v.size() - 1])))) ||
              ((v.size() > 0) && (v[v.size() - 1].fi == x) && (v[v.size() - 1].se >= y))) 
            v.pop_back(), sid[cur--] = 0;
        v.push_back(MP(x, y));  sid[++cur] = a[i].id;
    }
}
int cnt = 0;
db Find(int id, db k) {
    int tmp = v.size();
    int l = 0, r = v.size() - 1, ans = 0;
    while(l <= r) {
        int mid = l + r >> 1;
        if((mid == v.size() - 1) || (GetK(v[mid], v[mid + 1]) > k)) r = mid - 1, ans = mid;
        else l = mid + 1;
    }
    return v[ans].se - k * v[ans].fi;
}
void CDQ(int l, int r) {
    if(l == r) {
        a[l].Get(); 
        return ;   
    }
    int mid = l + r >> 1;
    CDQ(l, mid); 
    GetConvexHull(l, mid);
    for(int i = mid + 1; i <= r; i++) {
        chmin(a[i].f, Find(i, a[i].h) + a[i].ad);
    }
    CDQ(mid + 1, r);
    int tl = l, tr = mid + 1, tot = tl - 1;
    while(tl <= mid || tr <= r) {
        if((tr > r) || (tl <= mid && a[tl].x < a[tr].x)) st[++tot] = a[tl++];//?????tl <= mid 
        else st[++tot] = a[tr++];
    }
    for(int i = l; i <= r; i++) a[i] = st[i]; 
}
signed main() {
    N = read(); 
    for(int i = 1; i <= N; i++) a[i].h = read();
    for(int i = 1; i <= N; i++) {
        a[i].w = read() + a[i - 1].w; a[i].id = i;
        a[i].f = a[i - 1].f + sqr(a[i].h - a[i - 1].h);
        a[i].ad = sqr(a[i].h) + a[i - 1].w;
        if(i == 1) a[1].f = 0;
    }
    CDQ(1, N);
    sort(a + 1, a + N + 1, comp);
//  for(int i = 1; i <= N; i++) cout << i << ' ' << (LL)a[i].f << '\n';
    c