제출 #1222026

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1222026		395333em	Building Bridges (CEOI17_building)	C++20	100 / 100	29 ms	8712 KiB

building

#include <bits/stdc++.h>
#define int long long
#define emb emplace_back
#define pii pair <int, int>

using namespace std;

const int mod = 1e9 + 7;
const int inf = 1e18;
const int N = 1e5 + 5;
const int H = 1e6;

int n, h[N], w[N], pref[N], dp[N];

struct line {
    int m, c;
    int cal(int x){
        return m * x + c;
    }
    line(int m, int c) : m(m), c(c) {}
};

struct lichaotree {
    struct node {
        line val;
        node *l, *r;
        node(line val) : l(0), r(0), val(val) {}
    };
    typedef node* pnode;
    pnode root;
    void update(int l, int r, pnode &i, line curr){
        if (!i) return i = new node(curr), void();
        int mid = (l + r) / 2;
        if (curr.cal(mid) < i->val.cal(mid)) swap(curr, i->val);
        if (curr.cal(l) < i->val.cal(l)) update(l, mid, i->l, curr);
        if (curr.cal(r) < i->val.cal(r)) update(mid + 1, r, i->r, curr);
    }
    int query(int l, int r, pnode &i, int idx){
        if (!i) return inf;
        if (l == r) return i->val.cal(idx);
        int mid = (l + r) / 2;
        if (idx <= mid) return min(i->val.cal(idx), query(l, mid, i->l, idx));
        else return min(i->val.cal(idx), query(mid + 1, r, i->r, idx));
    }
} lct;

signed main(){
    cin.tie(NULL)->sync_with_stdio(false);
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> h[i];
    for (int i = 1; i <= n; i++) cin >> w[i], pref[i] = pref[i - 1] + w[i];
    dp[1] = 0;
    for (int i = 2; i <= n; i++) dp[i] = inf;
    lct.update(1, H, lct.root, line(-2 * h[1], dp[1] + h[1] * h[1] - pref[1]));
    for (int i = 2; i <= n; i++) {
        dp[i] = pref[i - 1] + h[i] * h[i] + lct.query(1, H, lct.root, h[i]);
        lct.update(1, H, lct.root, line(-2 * h[i], dp[i] + h[i] * h[i] - pref[i]));
    }
    cout << dp[n];
}

/*
dp[i] = min(dp[i], dp[j] + pref[i - 1] - pref[j] + (h[i] - h[j]) * (h[i] - h[j]))
= min(dp[i], dp[j] + pref[i - 1] - pref[j] + h[i]^2 - 2(h[i])(h[j]) + h[j]^2)
= pref[i - 1] + h[i]^2 + min_element(dp[j] - pref[j] - 2(h[i])(h[j]) + h[j]^2)
= pref[i - 1] + h[i]^2 + min_element(-2(h[i])(h[j]) + dp[j] - pref[j] + h[j]^2)
from y = mx + c
m = -2(h[j])
x = h[i]
c = dp[j] + h[j]^2 - pref[j]
*/

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