// // Monotone 単一始点最短路問題 by D&D Monotone Minima // 頂点数 N+1 の DAG, 頂点 i, j 間のコスト f(i, j) が Monotone であることを仮定 (argmin が単調非減少) // O(N (log N)^2) // // verified // AtCoder EDPC Z - Frog 3 // https://atcoder.jp/contests/dp/tasks/dp_z // // Codeforces Round 189 (Div. 1) C. Kalila and Dimna in the Logging Industry // https://codeforces.com/contest/319/problem/C // // yukicoder No.705 ゴミ拾い Hard // https://yukicoder.me/problems/no/705 // // Reference: // tatyam: Monge の手引き書 // https://speakerdeck.com/tatyam_prime/monge-noshou-yin-shu // #include using namespace std; // find min_j f(i, j) for all i, by Monotone Minima, O(H + W log H) // f(i, j) must be monotone (argmin is not decreasing) template void MonotoneMinimaRec (int HL, int HR, int WL, int WR, const FUNC &f, vector> &res) { if (HR - HL <= 0) return; int HM = (HL + HR) / 2; res[HM].second = WL; for (int i = WL; i < WR; i++) { VAL val = f(HM, i); if (res[HM].first > val) res[HM] = make_pair(val, i); } MonotoneMinimaRec(HL, HM, WL, res[HM].second + 1, f, res); MonotoneMinimaRec(HM + 1, HR, res[HM].second, WR, f, res); } template vector> MonotoneMinima(int H, int W, const FUNC &f) { vector> res(H, make_pair(numeric_limits::max() / 2, -1)); MonotoneMinimaRec(0, H, 0, W, f, res); return res; } // find shortest path on DAG with monotone cost, by D&D Monotone Minima, O(N (log N)^2) // vertex: 0, 1, 2, ..., N // f(i, j) must be monotone (argmin is not decreasing) template vector> MonotoneMinimaDD(int N, const FUNC &f) { vector> res(N + 1, make_pair(numeric_limits::max() / 2, -1)); res[0].first = VAL(0); auto f2 = [&](int i, int j) -> VAL { return res[j].first + f(j, i); }; auto rec = [&](auto &&rec, int left, int right) -> void { if (right - left <= 1) return; int mid = (left + right) / 2; rec(rec, left, mid); MonotoneMinimaRec(mid, right, left, mid, f2, res); //rec2(rec2, mid, right, left, mid); rec(rec, mid, right); }; rec(rec, 0, N + 1); return res; } //------------------------------// // Examples //------------------------------// // AtCoder EDPC Z - Frog 3 /* H は単調増加数列 chmin(dp[j], dp[i] + (H[j] - H[i])^2 + C) i -> j のコスト:(H[j] - H[i])^2 ...... 差の凸関数は Monge スタート: 0, ゴール: N-1 */ void EDPC_Z() { long long N, C; cin >> N >> C; vector H(N); for (long long i = 0; i < N; i++) cin >> H[i]; auto func = [&](int i, int j) -> long long { return (H[j] - H[i]) * (H[j] - H[i]) + C; }; auto res = MonotoneMinimaDD(N-1, func); cout << res[N-1].first << endl; } // Codeforces Round 189 (Div. 1) C. Kalila and Dimna in the Logging Industry /* A: 単調増加, B: 単調減少, ともに長さ N i -> j のコストが、B[i] × A[j] で与えられる ..... 単調増加 × 単調減少は Monge スタート: 0, ゴール: N-1 */ void Codeforces_189_C() { long long N; cin >> N; vector A(N), B(N); for (int i = 0; i < N; i++) cin >> A[i]; for (int i = 0; i < N; i++) cin >> B[i]; auto func = [&](int i, int j) -> long long { return B[i] * A[j]; }; auto res = MonotoneMinimaDD(N-1, func); cout << res[N-1].first << endl; } // yukicoder No.705 ゴミ拾い Hard /* A, X, Y: N 個 これらを区間に分割していく  dp[j] = min_{0 ≦ i < j}(dp[i] + |A[j-1] - X[i]|^3 + |-Y[i]|^3) i -> j のコスト:|A[j-1] - X[i]|^3 + |-Y[i]|^3 ...... 差の凸関数 (Monge) + 縞々 (Monge) -> Monge スタート: 0, ゴール: N */ void yukicoder_705() { int N; cin >> N; vector A(N), X(N), Y(N); for (int i = 0; i < N; i++) cin >> A[i]; for (int i = 0; i < N; i++) cin >> X[i]; for (int i = 0; i < N; i++) cin >> Y[i]; auto func = [&](int i, int j) -> long long { long long dx = abs(A[j-1] - X[i]), dy = abs(Y[i]); return dx * dx * dx + dy * dy * dy; }; auto res = MonotoneMinimaDD(N, func); cout << res[N].first << endl; } int main() { //EDPC_Z(); //Codeforces_189_C(); yukicoder_705(); }