제출 #995689

#제출 시각아이디문제언어결과실행 시간메모리
995689biankAliens (IOI16_aliens)C++17
0 / 100
0 ms348 KiB
#include<bits/stdc++.h> using namespace std; #define all(x) begin(x),end(x) #define sz(x) int(x.size()) using ll = long long; using ld = long double; struct Line { ll m, b, k; mutable ld p; //we need to change p in the set Line(ll _m, ll _b, ll _k) : m(_m), b(_b), k(_k), p(0) {} bool operator<(const Line& o) const { //order by slope return m < o.m; } bool operator<(ld x) const { return p < x; } ld intersectX(const Line l) const { return ld(b - l.b) / (l.m - m); } }; struct LineContainer : multiset<Line, less<>> { static constexpr ld INF = 1e18; bool intersect(iterator x, iterator y) { if (y == end()) { x->p = INF; return false; } if (x->m == y->m) { //parallel if (x->b > y->b) x->p = INF; else x->p = -INF; } else { x->p = x->intersectX(*y); } return x->p >= y->p; } void add(ll m, ll b, ll k) { auto next = emplace(m, b, k); auto prev = next, curr = next; next++; while (intersect(curr, next)) { next = erase(next); } if (prev != begin() && intersect(--prev, curr)) { curr = erase(curr); intersect(prev, curr); } curr = prev; while (curr != begin() && (--prev)->p >= curr->p) { curr = erase(curr); intersect(prev, curr); curr = prev; } } pair<ll, ll> query(ll x) { assert(!empty()); Line l = *lower_bound(x); return {l.m * x + l.b, l.k}; } }; struct Point { ll x, y; Point(ll _x, ll _y) : x(_x), y(_y) {} }; ll take_photos(int n, int /*m*/, int k, vector<int> r, vector<int> c){ vector<Point> v; for (int i = 0; i < n; i++) { if (r[i] < c[i]) swap(r[i], c[i]); v.emplace_back(r[i], c[i]); } sort(all(v), [](const Point &a, const Point &b) { if (a.y != b.y) return a.y < b.y; return a.x > b.x; }); vector<Point> p{v[0]}; for (int i = 1; i < n; i++) { if (p.back().x < v[i].x) p.push_back(v[i]); } n = sz(p); k = min(k, n); auto sq = [](ll x) { return x * x; }; auto check = [&](ll lambda) { LineContainer dp; ll res = 0, cnt = 0; for (int i = 0; i < n; i++) { ll m = -2LL * (p[i].y - 1LL); ll b = res + sq(p[i].y - 1LL); if (i > 0) b -= sq(max(p[i - 1].x - p[i].y + 1LL, 0LL)); dp.add(m, b, cnt); tie(res, cnt) = dp.query(p[i].x); res += sq(p[i].x) + lambda, cnt++; } return pair<ll, ll>{res, cnt}; }; ll lo = 0, hi = 1e13; while (hi - lo > 1) { ll mid = (hi - lo) / 2 + lo; if (check(mid).second <= k) hi = mid; else lo = mid; } return check(hi).first - k * hi; }
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