結果

問題 No.913 木の燃やし方
ユーザー はまやんはまやん
提出日時 2019-10-19 11:31:57
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 1,000 ms / 3,000 ms
コード長 6,318 bytes
コンパイル時間 2,756 ms
コンパイル使用メモリ 202,280 KB
最終ジャッジ日時 2025-01-07 23:59:49
ジャッジサーバーID
（参考情報） 		judge1 / judge1
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 * 3
other AC * 34
権限があれば一括ダウンロードができます
コンパイルメッセージ
main.cpp: In member function ‘bool ConvexHullDynamic::Line::operator<(const ConvexHullDynamic::Line&) const’:
main.cpp:50:17: warning: control reaches end of non-void function [-Wreturn-type]
   50 |                 }
      |                 ^

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>
#define rep(i,a,b) for(int i=a;i<b;i++)
#define rrep(i,a,b) for(int i=a;i>=b;i--)
#define fore(i,a) for(auto &i:a)
#define all(x) (x).begin(),(x).end()
//#pragma GCC optimize ("-O3")
using namespace std; void _main(); int main() { cin.tie(0); ios::sync_with_stdio(false); _main(); }
typedef long long ll; const int inf = INT_MAX / 2; const ll infl = 1LL << 60;
template<class T>bool chmax(T& a, const T& b) { if (a < b) { a = b; return 1; } return 0; }
template<class T>bool chmin(T& a, const T& b) { if (b < a) { a = b; return 1; } return 0; }
//---------------------------------------------------------------------------------------------------
class ConvexHullDynamic
{
    typedef long long coef_t;
    typedef long long coord_t;
    typedef long long val_t;
    /*
    * Line 'y=a*x+b' represented by 2 coefficients 'a' and 'b'
    * and 'xLeft' which is intersection with previous line in hull(first line has -INF)
    */
private:
    struct Line
    {
        coef_t a, b;
        double xLeft;
        enum Type
        {
            line, maxQuery, minQuery
        } type;
        coord_t val;
        explicit Line(coef_t aa = 0, coef_t bb = 0) : a(aa), b(bb), xLeft(-INFINITY), type(Type::line), val(0) {}
        val_t valueAt(coord_t x) const { return a * x + b; }
        friend bool areParallel(const Line& l1, const Line& l2) { return l1.a == l2.a; }
        friend double intersectX(const Line& l1, const Line& l2) { return areParallel(l1, l2) ? INFINITY : 1.0 * (l2.b - l1.b) / (l1.a - l2.a); }
        bool operator<(const Line& l2) const
        {
            if (l2.type == line)
                return this->a < l2.a;
            if (l2.type == maxQuery)
                return this->xLeft < l2.val;
            if (l2.type == minQuery)
                return this->xLeft > l2.val;
        }
    };
    bool isMax; //whether or not saved envelope is top(search of max value)
public:
    std::set< Line > hull;  //envelope itself
private:
    /*
    * INFO:        Check position in hull by iterator
    * COMPLEXITY:  O(1)
    */
    bool hasPrev(std::set< Line >::iterator it) { return it != hull.begin(); }
    bool hasNext(std::set< Line >::iterator it) { return it != hull.end() && std::next(it) != hull.end(); }
    /*
    * INFO:        Check whether line l2 is irrelevant
    * NOTE:        Following positioning in hull must be true
    *              l1 is next left to l2
    *              l2 is right between l1 and l3
    *              l3 is next right to l2
    * COMPLEXITY:  O(1)
    */
    bool irrelevant(const Line& l1, const Line& l2, const Line& l3) { return intersectX(l1, l3) <= intersectX(l1, l2); }
    bool irrelevant(std::set< Line >::iterator it)
    {
        return hasPrev(it) && hasNext(it)
            && (isMax && irrelevant(*std::prev(it), *it, *std::next(it))
                || !isMax && irrelevant(*std::next(it), *it, *std::prev(it)));
    }
    /*
    * INFO:        Updates 'xValue' of line pointed by iterator 'it'
    * COMPLEXITY:  O(1)
    */
    std::set< Line >::iterator updateLeftBorder(std::set< Line >::iterator it)
    {
        if (isMax && !hasPrev(it) || !isMax && !hasNext(it))
            return it;
        double val = intersectX(*it, isMax ? *std::prev(it) : *std::next(it));
        Line buf(*it);
        it = hull.erase(it);
        buf.xLeft = val;
        it = hull.insert(it, buf);
        return it;
    }
public:
    explicit ConvexHullDynamic(bool isMax = false) : isMax(isMax) {}
    /*
    * INFO:        Adding line to the envelope
    *              Line is of type 'y=a*x+b' represented by 2 coefficients 'a' and 'b'
    * COMPLEXITY:  Adding N lines(N calls of function) takes O(N*log N) time
    */
    void addLine(coef_t a, coef_t b)
    {
        //find the place where line will be inserted in set
        Line l3 = Line(a, b);
        auto it = hull.lower_bound(l3);
        //if parallel line is already in set, one of them becomes irrelevant
        if (it != hull.end() && areParallel(*it, l3)) {
            if (isMax && it->b < b || !isMax && it->b > b)
                it = hull.erase(it);
            else
                return;
        }
        //try to insert
        it = hull.insert(it, l3);
        if (irrelevant(it)) {
            hull.erase(it);
            return;
        }
        //remove lines which became irrelevant after inserting line
        while (hasPrev(it) && irrelevant(std::prev(it))) hull.erase(std::prev(it));
        while (hasNext(it) && irrelevant(std::next(it))) hull.erase(std::next(it));
        //refresh 'xLine'
        it = updateLeftBorder(it);
        if (hasPrev(it))
            updateLeftBorder(std::prev(it));
        if (hasNext(it))
            updateLeftBorder(std::next(it));
    }
    val_t getBest(coord_t x) const
    {
        Line q;
        q.val = x;
        q.type = isMax ? Line::Type::maxQuery : Line::Type::minQuery;
        auto bestLine = hull.lower_bound(q);
        if (isMax) --bestLine;
        return bestLine->valueAt(x);
    }
};
/*---------------------------------------------------------------------------------------------------
　　　　　　　　　　　 ∧＿∧
　　　　　 ∧＿∧ 　（´<_｀ ）　 Welcome to My Coding Space!
　　　　 （ ´_ゝ`）　/　 ⌒i     @hamayanhamayan
　　　　／　　　＼　 　  |　|
　　　 /　　 /￣￣￣￣/　　|
　 ＿_(__ﾆつ/　    ＿/ .| .|＿＿＿＿
　 　　　＼/＿＿＿＿/　（u　⊃
---------------------------------------------------------------------------------------------------*/
int N, A[201010];
ll B[201010];
//---------------------------------------------------------------------------------------------------
void f(int L, int R, vector<ll>& ans, bool isRev = false) {
    if (L + 1 == R) {
        chmin(ans[L], 1LL + A[L]);
        return;
    }
    int md = (L + R) / 2;
    if (isRev and (R - L) % 2 == 1) md++;
    f(L, md, ans, isRev);
    f(md, R, ans, isRev);
    ConvexHullDynamic cht;
    rep(i, md, R) cht.addLine(-2LL * (i + 1), 1LL * (i + 1) * (i + 1) + B[i]);
    
    ll mi = infl;
    rep(i, L, md) {
        ll cst = cht.getBest(i) + 1LL * i * i;
        if (i) cst -= B[i - 1];
        chmin(mi, cst);
        chmin(ans[i], mi);
    }
}
//---------------------------------------------------------------------------------------------------
void _main() {
    cin >> N;
    rep(i, 0, N) cin >> A[i];
    vector<ll> normal(N, infl);
    B[0] = A[0];
    rep(i, 1, N) B[i] = B[i - 1] + A[i];
    f(0, N, normal);
    reverse(A, A + N);
    vector<ll> rev(N, infl);
    B[0] = A[0];
    rep(i, 1, N) B[i] = B[i - 1] + A[i];
    f(0, N, rev, true);
    rep(i, 0, N) printf("%lld\n", min(normal[i], rev[N - 1 - i]));
}
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