loj#2483. 「CEOI2017」Building Bridges(dp cdq 凸包)

题意

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';
    cout << (ll)a[n].f;
    return 0;
}