제출 #902602

#제출 시각아이디문제언어결과실행 시간메모리
902602joelgun14Building Bridges (CEOI17_building)C++17
100 / 100
107 ms13048 KiB
// header file #include <bits/stdc++.h> // pragma #pragma GCC optimize("Ofast") #pragma GCC optimize("unroll-loops") // macros #define endl "\n" #define ll long long #define mp make_pair #define ins insert #define lb lower_bound #define pb push_back #define ub upper_bound #define lll __int128 #define fi first #define se second using namespace std; ll sqr(ll a) { return a * a; } const ll is_query = -(1LL<<62); struct line { ll m, b; mutable function<const line*()> succ; bool operator<(const line &rhs) const { if(rhs.b != is_query) return m < rhs.m; const line* s = succ(); if(!s) return 0; ll x = rhs.m; return b - s -> b < (s -> m - m) * x; } }; struct dynamic_hull : public multiset<line> { const ll inf = LLONG_MAX; bool bad(iterator y) { auto z = next(y); // basically if the line is the same // if the line is the same then we will pick the lower line if(y == begin()) { if(z == end()) return 0; return y -> m == z -> m && y -> b <= z -> b; } auto x = prev(y); if(z == end()) { return y -> m == x -> m && y -> b <= x -> b; } // v1 is with prev ll v1 = (x -> b - y -> b); // check in case slope is same if(y -> m == x -> m) v1 = x -> b > y -> b ? inf : -inf; else v1 /= (y -> m - x -> m); // v2 is with next ll v2 = (y -> b - z -> b); if(z -> m == y -> m) v2 = y -> b > z -> b ? inf : -inf; else v2 /= (z -> m - y -> m); return v1 >= v2; } void insert_line(ll m, ll b) { auto y = insert({m, b}); y -> succ = [=] {return next (y) == end() ? 0 : &*next(y); }; if(bad(y)) { erase(y); return; } while(next(y) != end() && bad(next(y))) erase(next(y)); while(y != begin() && bad(prev(y))) erase(prev(y)); } ll eval(ll x) { auto l = *lower_bound((line) {x, is_query}); return l.m * x + l.b; } }; dynamic_hull cht; int main() { ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0); int n; cin >> n; int h[n], w[n]; for(int i = 0; i < n; ++i) cin >> h[i]; for(int i = 0; i < n; ++i) cin >> w[i]; // we choose first and we choose last // the problem is we need to choose some number of pillars // the factors of choosing a pillar // height // cost of removing the pillar // use dynamic cht dp opt ll dp[n], pref[n]; memset(dp, 0, sizeof(dp)); pref[0] = 0; for(int i = 1; i < n; ++i) pref[i] = pref[i - 1] + w[i]; cht.insert_line(2 * h[0], -sqr(h[0])); for(int i = 1; i < n; ++i) { dp[i] = -cht.eval(h[i]) + pref[i - 1] + sqr(h[i]); cht.insert_line(2 * h[i], -(dp[i] + sqr(h[i]) - pref[i])); } cout << dp[n - 1] << endl; return 0; }
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