欢迎您访问程序员文章站本站旨在为大家提供分享程序员计算机编程知识!
您现在的位置是: 首页  >  IT编程

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

程序员文章站 2022-06-10 17:18:30
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~~ cpp include define Pair pair defi ......

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;
}