#ifndef HIDDEN_IN_VS // 折りたたみ用 // 警告の抑制 #define _CRT_SECURE_NO_WARNINGS // ライブラリの読み込み #include using namespace std; // 型名の短縮 using ll = long long; using ull = unsigned long long; // -2^63 ~ 2^63 = 9e18(int は -2^31 ~ 2^31 = 2e9) using pii = pair; using pll = pair; using pil = pair; using pli = pair; using vi = vector; using vvi = vector; using vvvi = vector; using vvvvi = vector; using vl = vector; using vvl = vector; using vvvl = vector; using vvvvl = vector; using vb = vector; using vvb = vector; using vvvb = vector; using vc = vector; using vvc = vector; using vvvc = vector; using vd = vector; using vvd = vector; using vvvd = vector; template using priority_queue_rev = priority_queue, greater>; using Graph = vvi; // 定数の定義 const double PI = acos(-1); int DX[4] = { 1, 0, -1, 0 }; // 4 近傍(下,右,上,左) int DY[4] = { 0, 1, 0, -1 }; int INF = 1001001001; ll INFL = 4004004003094073385LL; // (int)INFL = INF, (int)(-INFL) = -INF; // 入出力高速化 struct fast_io { fast_io() { cin.tie(nullptr); ios::sync_with_stdio(false); cout << fixed << setprecision(18); } } fastIOtmp; // 汎用マクロの定義 #define all(a) (a).begin(), (a).end() #define sz(x) ((int)(x).size()) #define lbpos(a, x) (int)distance((a).begin(), std::lower_bound(all(a), (x))) #define ubpos(a, x) (int)distance((a).begin(), std::upper_bound(all(a), (x))) #define Yes(b) {cout << ((b) ? "Yes\n" : "No\n");} #define rep(i, n) for(int i = 0, i##_len = int(n); i < i##_len; ++i) // 0 から n-1 まで昇順 #define repi(i, s, t) for(int i = int(s), i##_end = int(t); i <= i##_end; ++i) // s から t まで昇順 #define repir(i, s, t) for(int i = int(s), i##_end = int(t); i >= i##_end; --i) // s から t まで降順 #define repe(v, a) for(const auto& v : (a)) // a の全要素(変更不可能) #define repea(v, a) for(auto& v : (a)) // a の全要素(変更可能) #define repb(set, d) for(int set = 0, set##_ub = 1 << int(d); set < set##_ub; ++set) // d ビット全探索(昇順) #define repis(i, set) for(int i = lsb(set), bset##i = set; i < 32; bset##i -= 1 << i, i = lsb(bset##i)) // set の全要素(昇順) #define repp(a) sort(all(a)); for(bool a##_perm = true; a##_perm; a##_perm = next_permutation(all(a))) // a の順列全て(昇順) #define uniq(a) {sort(all(a)); (a).erase(unique(all(a)), (a).end());} // 重複除去 #define EXIT(a) {cout << (a) << endl; exit(0);} // 強制終了 #define inQ(x, y, u, l, d, r) ((u) <= (x) && (l) <= (y) && (x) < (d) && (y) < (r)) // 半開矩形内判定 // 汎用関数の定義 template inline ll powi(T n, int k) { ll v = 1; rep(i, k) v *= n; return v; } template inline bool chmax(T& M, const T& x) { if (M < x) { M = x; return true; } return false; } // 最大値を更新(更新されたら true を返す) template inline bool chmin(T& m, const T& x) { if (m > x) { m = x; return true; } return false; } // 最小値を更新(更新されたら true を返す) template inline T getb(T set, int i) { return (set >> i) & T(1); } template inline T smod(T n, T m) { n %= m; if (n < 0) n += m; return n; } // 非負mod // 演算子オーバーロード template inline istream& operator>>(istream& is, pair& p) { is >> p.first >> p.second; return is; } template inline istream& operator>>(istream& is, vector& v) { repea(x, v) is >> x; return is; } template inline vector& operator--(vector& v) { repea(x, v) --x; return v; } template inline vector& operator++(vector& v) { repea(x, v) ++x; return v; } #endif // 折りたたみ用 #if __has_include() #include using namespace atcoder; #ifdef _MSC_VER #include "localACL.hpp" #endif using mint = modint998244353; //using mint = static_modint<(int)1e9 + 7>; //using mint = modint; // mint::set_mod(m); namespace atcoder { inline istream& operator>>(istream& is, mint& x) { ll x_; is >> x_; x = x_; return is; } inline ostream& operator<<(ostream& os, const mint& x) { os << x.val(); return os; } } using vm = vector; using vvm = vector; using vvvm = vector; using vvvvm = vector; using pim = pair; #endif #ifdef _MSC_VER // 手元環境(Visual Studio) #include "local.hpp" #else // 提出用(gcc) inline int popcount(int n) { return __builtin_popcount(n); } inline int popcount(ll n) { return __builtin_popcountll(n); } inline int lsb(int n) { return n != 0 ? __builtin_ctz(n) : 32; } inline int lsb(ll n) { return n != 0 ? __builtin_ctzll(n) : 64; } inline int msb(int n) { return n != 0 ? (31 - __builtin_clz(n)) : -1; } inline int msb(ll n) { return n != 0 ? (63 - __builtin_clzll(n)) : -1; } #define dump(...) #define dumpel(v) #define dump_math(v) #define input_from_file(f) #define output_to_file(f) #define Assert(b) { if (!(b)) { vc MLE(1<<30); EXIT(MLE.back()); } } // RE の代わりに MLE を出す #endif //【正規括弧列 → 木】O(n) /* * 正規括弧列 s[0..2n) について,ネスト関係を表した 0 を根とする有向根付き木 g[0..n] を返す. * i 番目の頂点は対応する括弧の組 s[ls[i]] = '(', s[rs[i]] = ')' に対応し,子ほどネストが深いものとする. * ただし ls[0] = -1, rs[0] = 2n とする. */ Graph parenthesis_tree(const string& s, vi* ls = nullptr, vi* rs = nullptr) { // verify : https://atcoder.jp/contests/discovery2016-final/tasks/discovery_2016_final_c int n = sz(s) / 2; Graph g(n + 1); if (ls) ls->resize(n + 1); if (rs) rs->resize(n + 1); int id = 1; stack stk; // ('(' の位置, 木の頂点番号) stk.push({ -1, 0 }); if (ls) (*ls)[0] = -1; if (rs) (*rs)[0] = 2 * n; rep(i, 2 * n) { if (s[i] == '(') { stk.push({ i, id++ }); } else { auto [l, v] = stk.top(); stk.pop(); g[stk.top().second].push_back(v); if (ls) (*ls)[v] = l; if (rs) (*rs)[v] = i; } } return g; } // 貪欲は根からも葉からもダメ void WA() { int n; string s; cin >> n >> s; vi a(n); cin >> a; vi ls, rs; auto g = parenthesis_tree(s, &ls, &rs); dump(ls); dump(rs); vi acc(2 * n + 1); rep(i, 2 * n) acc[i + 1] = acc[i] + (s[i] == '('); ll res = 0; priority_queue q; ll w = 1; q.push({ 0, 0 }); while (!q.empty()) { auto [v, i] = q.top(); q.pop(); if (i == 0) { repe(c, g[i]) { int v2 = a[acc[ls[c]]]; q.push({ v2, c }); } } else { res += w * v; repe(c, g[i]) { int v2 = a[acc[ls[c]]]; q.push({ v2, c }); } w++; } } // 計算量おかしいけどノーペナなのでとりあえず提出 EXIT(res); } pair solve(int n, vi par, vl c0, vl c1) { dsu d(n); using S = tuple; auto comp = [&](const S& l, const S& r) { auto [c0_l, c1_l, id_l] = l; auto [c0_r, c1_r, id_r] = r; // (0, 0) は最小ってことにする. if (c0_r == 0 && c1_r == 0) return false; if (c0_l == 0 && c1_l == 0) return true; // キューから昇順に取り出したいので不等号の向きは逆にする. return c0_r * c1_l > c0_l * c1_r; }; priority_queue, decltype(comp)> q(comp); rep(i, n) q.push({ c0[i], c1[i], i }); // ソートした列も得る(マージテク) vector> seq(n); rep(i, n) seq[i].push_back(i); ll res = 0; while (!q.empty()) { dump("------------------------------------"); auto [c0_s, c1_s, id_s] = q.top(); q.pop(); id_s = d.leader(id_s); int id_p = par[id_s]; if (id_p == -1 || c0_s != c0[id_s] || c1_s != c1[id_s]) continue; id_p = d.leader(id_p); res += c1[id_p] * c0_s; d.merge(id_p, id_s); auto ld = d.leader(id_p); if (sz(seq[id_p]) > sz(seq[id_s])) { while (!seq[id_s].empty()) { seq[id_p].push_back(seq[id_s].front()); seq[id_s].pop_front(); } if (id_p != ld) swap(seq[id_p], seq[id_s]); } else { while (!seq[id_p].empty()) { seq[id_s].push_front(seq[id_p].back()); seq[id_p].pop_back(); } if (id_s != ld) swap(seq[id_p], seq[id_s]); } par[ld] = par[id_p]; c0[ld] = c0[id_p] + c0_s; c1[ld] = c1[id_p] + c1_s; q.push({ c0[ld], c1[ld], ld }); dump(id_s); dump(d); dump(q); dump(c0); dump(c1); dumpel(seq); } vi p(n); int ld = d.leader(0); rep(i, n) p[i] = seq[ld][i]; return { res, p }; } int main() { // input_from_file("input.txt"); // output_to_file("output.txt"); int n; string s; cin >> n >> s; vi a(n); cin >> a; vi ls, rs; auto g = parenthesis_tree(s, &ls, &rs); dump(ls); dump(rs); vi acc(2 * n + 1); rep(i, 2 * n) acc[i + 1] = acc[i] + (s[i] == '('); vi par(n + 1); par[0] = -1; rep(s, n + 1) repe(t, g[s]) par[t] = s; vl c0(n + 1); c0[0] = 0; repi(i, 1, n) c0[i] = a[acc[ls[i]]]; vl c1(n + 1, 1); dump(par); dump(c0); dump(c1); auto [res, p] = solve(n + 1, par, c0, c1); // https://judge.yosupo.jp/problem/rooted_tree_topological_order_with_minimum_inversions EXIT(res); }