Submission #503735

#	Time	Username	Problem	Language	Result	Execution time	Memory
503735		fvogel499	Building Bridges (CEOI17_building)	C++17	100 / 100	51 ms	36664 KiB

building

/*
 * File created on 01/08/2022 at 19:13:29.
 * Link to problem: 
 * Description: 
 * Time complexity: O()
 * Space complexity: O()
 * Status: ---
 * Copyright: Ⓒ 2022 Francois Vogel
*/

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <functional>

using namespace std;
using namespace __gnu_pbds;

#define pii pair<int, int>
#define f first
#define s second

#define vi vector<int>
#define all(x) x.begin(), x.end()
#define size(x) (int)((x).size())
#define pb push_back
#define ins insert
#define cls clear

#define int ll
#define ll long long
#define ld long double

typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;

const int p2 = 1<<20;
const int inf = 1e18;

struct LiChao {
    LiChao() {
        t = new pii [p2*2];
        for (int i = 0; i < p2*2; i++) t[i] = pii(0, inf);
    }
    void update(pii v, int x = 1, int b = 0, int e = p2-1) {
        int m = (b+e)/2;
        if (t[x].f*m+t[x].s > v.f*m+v.s) swap(t[x], v);
        if (b < e) {
            if (v.f > t[x].f) {
                update(v, x*2, b, m);
            }
            else {
                update(v, x*2+1, m+1, e);
            }
        }
    }
    int get(int x) {
        int r = inf;
        for (int i = p2+x; i; i /= 2) r = min(r, t[i].f*x+t[i].s);
        return r;
    }
    pii* t;
};

signed main() {
    cin.tie(0);
    ios_base::sync_with_stdio(0);

    int n;
    cin >> n;
    int b [n];
    for (int i = 0; i < n; i++) cin >> b[i];
    int w [n];
    for (int i = 0; i < n; i++) cin >> w[i];
    LiChao lc = LiChao();
    int dp [n];
    for (int i = n-1; i >= 0; i--) {
        dp[i] = lc.get(b[i]);
        if (i == n-1) dp[i] = 0;
        dp[i] += 2*b[i]*b[i];
        dp[i] -= w[i];
        lc.update(pii(-2*b[i], dp[i]));
    }
    int res = dp[0]-b[0]*b[0]-b[n-1]*b[n-1];
    for (int i : w) res += i;
    cout << res;

    cout.flush();
    int d = 0;
    d++;
}

• Subtask 1: 30 / 30
• Subtask 2: 30 / 30
• Subtask 3: 40 / 40