結果

問題 No.705 ゴミ拾い Hard
ユーザー anta
提出日時 2018-06-16 02:09:09
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 281 ms / 1,500 ms
コード長 3,058 bytes
コンパイル時間 1,943 ms
コンパイル使用メモリ 180,740 KB
実行使用メモリ 14,912 KB
最終ジャッジ日時 2024-06-30 15:55:48
合計ジャッジ時間 8,910 ms
ジャッジサーバーID
（参考情報） 		judge1 / judge3
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 4
other AC * 40
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

#include "bits/stdc++.h"
using namespace std;
template<typename T, typename U> static void amin(T &x, U y) { if (y < x) x = y; }
const long long INFL = 0x3f3f3f3f3f3f3f3fLL;
int main() {
    int N;
    while (~scanf("%d", &N)) {
        vector<int> as(N);
        for (int i = 0; i < N; ++ i)
            scanf("%d", &as[i]);
        vector<int> xs(N);
        for (int i = 0; i < N; ++ i)
            scanf("%d", &xs[i]);
        vector<int> ys(N);
        for (int i = 0; i < N; ++ i)
            scanf("%d", &ys[i]);
        sort(as.begin(), as.end());
        sort(xs.begin(), xs.end());
        auto values = xs;
        sort(values.begin(), values.end());
        values.erase(unique(values.begin(), values.end()), values.end());
        auto cb = [](int x) { return (long long)x * x * x; };
        auto monotoneMinima = [](int n, int m, auto f, auto out) {
            auto rec = [=](auto rec, int L, int R, int jL, int jR) -> void {
                if (L > R) return;
                int mid = (L + R) / 2;
                auto t = f(mid, jL);
                int jM = jL;
                for (int j = jL + 1; j <= jR; ++ j) {
                    auto u = f(mid, j);
                    if (u < t) t = u, jM = j;
                }
                out(mid, t, jM);
                rec(rec, L, mid - 1, jL, jM);
                rec(rec, mid + 1, R, jM, jR);
            };
            rec(rec, 0, n - 1, 0, m - 1);
        };
        auto totallyMonotoneMinima = [](int N, int M, auto f, auto out) {
            vector<int> minima(N, -1);
            auto rec = [N, f, out, &minima](auto rec, int k, int L, const vector<int> &js) -> void {
                if (L >= N) return;
                int s = 1 << k, n = ((N - L - 1) >> k) + 1;
                vector<int> njs;
                for (int j : js) {
                    int i = L + ((int)njs.size() << k);
                    while (!njs.empty() && !(f(i, njs.back()) < f(i, j)))
                        njs.pop_back(), i -= s;
                    if ((int)njs.size() + 1 < n)
                        njs.push_back(j);
                }
                rec(rec, k + 1, L + s, njs);
                for (int i = L, jx = 0; i < N; i += s * 2) {
                    int jR = i + s < N ? minima[i + s] : js.back();
                    auto t = f(i, js[jx]); int minj = js[jx];
                    for (++ jx; jx != js.size() && js[jx] <= jR; ++ jx) {
                        auto u = f(i, js[jx]);
                        if (u < t) t = u, minj = js[jx];
                    }
                    minima[i] = minj;
                    out(i, t, minj);
                    if (js[jx - 1] == jR) -- jx;
                }
            };
            vector<int> js;
            for (int j = 0; j < M; ++ j) js.push_back(j);
            rec(rec, 0, 0, js);
            assert(count(minima.begin(), minima.end(), -1) == 0);
        };
        auto solveOffline = [totallyMonotoneMinima](const long long *prev, int m, int n, auto f, auto out) {
            totallyMonotoneMinima(n, m, [=](int i, int j) {
                return prev[j] + f(j, i);
            }, [=](int i, long long t, int minj) {
                out(i, t);
            });
        };
        vector<long long> dp(N + 1, INFL);
        dp[0] = 0;
        vector<long long> tmp(N + 1);
        function<void(int, int)> rec = [&](int L, int R) {
            if (R - L <= 1) return;
            int mid = (L + R) / 2;
            rec(L, mid);
            for (int i = 0; i < R - L; ++ i)
                tmp[i] = dp[L + i] + (L + i == N ? 0 : cb(ys[L + i]));
            solveOffline(tmp.data(), mid - L, R - mid, [&](int j, int i) {
                j += L, i += mid;
                return cb(abs(as[i - 1] - xs[j]));
            }, [&](int i, long long t) {
                i += mid;
                amin(dp[i], t);
            });
            rec(mid, R);
        };
        rec(0, N + 1);
        long long ans = dp[N];
        printf("%lld\n", ans);
    }
}
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