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QOJ

IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#723700#9520. Concave Hullnan01TL 71ms9384kbC++2314.8kb2024-11-07 23:47:292024-11-07 23:47:29

Judging History

你现在查看的是最新测评结果

  • [2024-11-07 23:47:29]
  • 评测
  • 测评结果:TL
  • 用时:71ms
  • 内存:9384kb
  • [2024-11-07 23:47:29]
  • 提交

answer

#include <bits/stdc++.h>
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define per(i, a, n) for (int i = n; i >= a; i--)
#define ll long long
#define pii pair<int, int>
#define inf 0x3f3f3f3f
#define fi first
#define pll pair<ll, ll>
#define se second
#define lll __int128
#define ld long double
#define all(a) a.begin() + 1, a.end()
#define iofast                   \
    ios::sync_with_stdio(false); \
    cin.tie(0), cout.tie(0)
#define lowbit(i) (i & (-i))
using namespace std;
const int maxn = 1e6 + 5, mod = 1000000007;
// 三角形重心是 (x1+ x2+x3)/3, (y1+y2 + y3)/3
// if 外心= (0,0)
// -> 垂心=(x1+x2+x3, y1+y2+y3) (欧拉线:重心,外心,垂心共线,2*dis(外心,重心)=dis(重心,垂心)
// 内心=角平分线交点,角平分线与外接圆的三个交点组成的三角形的垂心等于原三角形的内心
const double eps = 1e-10;
int sgn(double a)
{
    if (fabs(a) < eps)
        return 0;
    if (a < 0)
        return -1;
    return 1;
}
template <class T>
class point
{
public:
    T x, y;
    point(T a, T b) : x(a), y(b) {}
    point() {};
    point operator+(point a) { return point(x + a.x, y + a.y); }
    point operator-(point a) { return point(x - a.x, y - a.y); }
    point operator*(double a) { return point(x * a, y * a); }
    point operator/(double a) { return point(x / a, y / a); }
    T operator*(point a) { return x * a.x + y * a.y; }
    T operator%(point a) { return x * a.y - y * a.x; }
    bool operator==(point a) { return x == a.x && y == a.y; }
    friend istream &operator>>(istream &is, point<T> &x)
    {
        is >> x.x >> x.y;
        return is;
    }
    friend ostream &operator<<(ostream &os, const point<T> &x) { return os << x.x << " " << x.y; }
};
template <class T>
struct circle
{
    point<T> o;
    T r;
};
template <class T>
class line
{
public:
    point<T> s, t;
};
template <typename T>
T mul(point<T> a, point<T> b, point<T> c) // c和ab叉乘 当ca逆时针转小于180度遇到cb时为正
{
    return (c - a) % (c - b);
}
template <typename T>
T dot(point<T> a, point<T> b, point<T> c) // ab和ac点乘
{
    return (a - b) * (a - c);
}
template <typename T>
point<T> interpoint(point<T> s1, point<T> t1, point<T> s2, point<T> t2) // 向量交点
{
    auto t = fabs(((t2 - s2) % (s2 - s1)) / ((t2 - s2) % (t1 - s1)));
    return s1 + (t1 - s1) * fabs(((t2 - s2) % (s2 - s1)) / ((t2 - s2) % (t1 - s1)));
}
template <typename T>
point<T> cross(point<T> a, point<T> b, point<T> c, point<T> d) // 直线交点
{
    auto v = c - d;
    ld sa = v % (a - d), sb = (b - d) % v;
    ld rate = sa / (sa + sb);
    return (b - a) * rate + a;
}
template <typename T>
inline bool parallel(point<T> s1, point<T> t1, point<T> s2, point<T> t2) // 判断平行 最好用整数point,别用double
{
    return ((t1 - s1) % (t2 - s2)) == 0;
}
template <typename T>
inline bool isinseg(point<T> s, point<T> t, point<T> mid) // 判断mid是否在st线段内,double改成eps
{
    return dot(mid, s, t) <= 0;
}
template <typename T>
double dis(point<T> a, point<T> b) // ab距离
{
    return sqrt((a - b) * (a - b));
}
template <typename T>
T distoline(point<T> a, point<T> b, point<T> c) // c到ab直线距离
{
    auto u = a - b, v = a - c;
    return fabs(u % v / dis(a, b));
}
template <typename T>
T distoseg(point<T> a, point<T> b, point<T> c) // c到ab线段距离
{
    auto u = b - a, v = c - a, w = c - b;
    if (a == b)
        return dis(c, b);
    if (u * v < 0)
        return dis(c, a);
    if (u * w > 0)
        return dis(b, c);
    return distoline(a, b, c);
}
template <typename T>
T sqadis(point<T> a, point<T> b) // ab平方距离
{
    return (a - b) * (a - b);
}
template <typename T>
bool cmp(point<T> a, point<T> b) // x第一关键字,y第二关键字
{
    return a.x == b.x ? a.y < b.y : a.x < b.x;
}
template <typename T>
point<T> rot(point<T> a) // 逆时针旋转90
{
    return point{-a.y, a.x};
}
template <typename T>
point<T> rotate(point<T> a, double angle) // 顺时针旋转angle度
{
    return point{a.x * cos(angle) + a.y * sin(angle), -a.x * sin(angle) + a.y * cos(angle)};
}
template <typename T>
double angle(point<T> a, point<T> b, point<T> c) // 返回b-a和c-a的tan值 垂直则返回-1;
{
    auto v1 = b - a;
    auto v2 = c - a;
    if (v1 * v2 == 0)
        return -1;
    return double(v1 % v2) / (v1 * v2);
}
template <typename T>
point<T> excir(point<T> a, point<T> b, point<T> c) // 求外接圆圆心
{
    auto d1 = (a + b) * 0.5, d2 = d1 + rot(a - b);
    auto d3 = (c + b) * 0.5, d4 = d3 + rot(c - b);
    return cross(d1, d2, d3, d4);
}
template <typename T>
bool check(circle<T> a, point<T> b) // 判断点是否在圆外
{
    if (dis(a.o, b) - a.r > eps)
        return false;
    else
        return 1;
}
template <typename T>
circle<T> smallestcircle(vector<point<T>> a) // 最小覆盖圆
{
    int n = a.size() - 1;
    random_shuffle(a.begin() + 1, a.begin() + n + 1);
    circle<T> c{a[1], 0};
    rep(i, 2, n)
    {
        if (!check(c, a[i]))
        {
            c = circle<T>{a[i], 0};
            rep(j, 1, i - 1)
            {
                if (!check(c, a[j]))
                {
                    c = circle<T>{(a[i] + a[j]) * 0.5, dis(a[i], a[j]) / 2.0};
                    rep(k, 1, j - 1)
                    {
                        if (!check(c, a[k]))
                        {
                            point<double> o = excir(a[i], a[j], a[k]);
                            c = circle<T>{o, dis(o, a[i])};
                        }
                    }
                }
            }
        }
    }
    return c;
}
template <typename T>
bool isinter(point<T> a, point<T> b, point<T> c, point<T> d) // 两线段是否相交
{
    double c1 = (c - a) % (b - a), c2 = (d - a) % (b - a);
    double c3 = (b - d) % (c - d), c4 = (a - d) % (c - d);
    return sgn(c1) * sgn(c2) < 0 && sgn(c3) * sgn(c4) < 0;
}
template <typename T>
vector<point<T>> graham(vector<point<T>> p, bool flag = 1) // 求凸包  1-index
{
    sort(all(p), cmp<T>);
    int n = p.size() - 1;
    if (n == 1)
        return p;
    vector<point<T>> h(n + 2);
    h[1] = p[1];
    int m = 1;
    rep(i, 2, n)
    {
        while (m > 1 && mul(h[m - 1], h[m], p[i]) <= 0)
            m--;
        h[++m] = p[i];
    }
    int k = m;
    for (int i = n - 1; i >= 1; i--)
    {
        while (m > k && mul(h[m - 1], h[m], p[i]) <= 0)
            m--;
        h[++m] = p[i];
    }
    h.resize(m + flag); // flag=1 不删除最后一个点 反之删除  最后一个点也是第一个点
    return h;
}
template <typename T>
bool ispointinhull(vector<point<T>> p, point<T> a) // log判断点是否再凸包内
{
    int n = p.size() - 1;
    int l = 0, r = n;
    while (l < r - 1)
    {
        int mid = l + r >> 1;
        T x = mul(p[mid], a, p[0]);
        T y = mul(p[mid + 1], a, p[0]);
        if (sgn(x) >= 0 && sgn(y) <= 0)
        {
            if (sgn(mul(p[mid + 1], a, p[mid])) >= 0)
            {
                return 1;
            }
            return 0;
        }
        else if (sgn(y) > 0)
            l = mid;
        else
            r = mid;
    }
    return 0;
}
template <typename T>
double zuijindiandui(vector<point<T>> a) // 最近公共点对
{
    sort(all(a), cmp<T>);
    double mn = 2e18;
    int n = a.size() - 1;
    auto cal = [&](auto self, int l, int r) -> void
    {
        if (l == r)
            return;
        int m = l + r >> 1;
        self(self, l, m);
        self(self, m + 1, r);
        double mid = (a[m].x + a[m + 1].x) / 2;
        int L = m;
        while (L >= l && a[L].x >= mid - mn)
        {
            L--;
        }
        L++;
        int R = m + 1;
        while (R <= r && a[R].x <= mid + mn)
            R++;
        R--;
        rep(i, L, m)
        {
            rep(j, m + 1, R)
            {
                mn = min(mn, dis(a[i], a[j]));
            }
        }
    };
    cal(cal, 1, n);
    return mn;
}
template <typename T>
T xzkk(vector<point<T>> a) // 旋转卡壳
{
    a = graham(a, 0); // 要删掉最后一个点!
    T mx = 0;
    int j = 3;
    int n = a.size() - 1;
    if (n == 2) // 如果是直线
    {
        return dis(a[1], a[2]);
    }
    a.push_back(a[1]);
    rep(i, 1, n)
    {
        mx = max(mx, dis(a[i], a[i + 1]));
        while (llabs(mul(a[i], a[i + 1], a[j])) <= llabs(mul(a[i], a[i + 1], a[j + 1])))
            j = (j % n) + 1;
        mx = max(mx, max(dis(a[i], a[j]), dis(a[i + 1], a[j])));
    }

    return mx;
}
template <typename T>
double angle(line<T> a) // 直线的极角
{
    return atan2(a.t.y - a.s.y, a.t.x - a.s.x);
}
template <typename T>
inline point<T> cross(line<T> a, line<T> b) // 两直线交点
{
    return cross(a.s, a.t, b.s, b.t);
}
template <typename T>
bool right(line<T> a, line<T> b, line<T> c) // bc交点时候在a右边
{
    auto d = cross(b, c);
    return mul(a.s, a.t, d) < 0; // 在线上不算
}
template <typename T>
bool left(line<T> a, line<T> b, line<T> c) // bc交点时候在a右边
{
    auto d = cross(b, c);
    return mul(a.s, a.t, d) >= 0; // 在线上算
}

template <typename T>
vector<point<T>> halfplane(vector<line<T>> a) // 半平面交
{
    sort(all(a), [&](auto a, auto b)
         { double x = angle(a), y = angle(b);
        return fabs(x-y) > eps ? x<y : mul(b.s, b.t, a.t) > 0; });
    deque<line<T>> q;
    q.push_back(a[1]);
    rep(i, 2, (int)a.size() - 1)
    {
        if (fabs(angle(a[i]) - angle(q.back()) < eps))
            continue;
        while (q.size() > 1)
        {
            auto t = q.back();
            q.pop_back();
            if (left(a[i], t, q.back()))
            {
                q.push_back(t);
                break;
            }
        }
        while (q.size() > 1)
        {
            auto t = q.front();
            q.pop_front();
            if (left(a[i], t, q.front()))
            {
                q.push_front(t);
                break;
            }
        }
        q.push_back(a[i]);
    }
    while (q.size() > 2)
    {
        auto t = q.back();
        q.pop_back();
        if (left(q.front(), t, q.back()))
        {
            q.push_back(t);
            break;
        }
    }
    int siz = q.size();
    q.push_back(q.front());
    vector<point<T>> ret(1); // 1-index
    rep(i, 1, siz)
    {
        auto t = q.front();
        q.pop_front();
        ret.push_back(cross(t, q.front()));
    }
    return ret;
}
template <typename T>
ll area(vector<point<T>> a) // 一个凸包求面积
{
    int siz = a.size() - 1;
    ll ret = 0;
    rep(i, 2, siz - 1)
    {
        ret += mul(a[1], a[i], a[i + 1]);
    }
    return ret;
}
template <typename T>
struct matrix
{
    point<T> a, b, c, d;
    T area;
    matrix(point<T> a, point<T> b, point<T> c, point<T> d, T area = 0) : a(a), b(b), c(c), d(d), area(area) {}
    matrix() : area(0) {}
};
template <typename T>
matrix<T> minmatcover(vector<point<T>> p) // p为1-index  最小矩形覆盖 (最小正方形覆盖三分)
{
    p = graham(p);
    int n = (int)p.size() - 2;
    p[0] = p[n];
    // j右边 k上边 h左边
    double ret = 2e18;
    array<int, 4> ansi;
    for (int i = 1, j = 1, k = 2, h = -1; i <= n; i++)
    {
        // cout << p[i].x << " " << p[i].y << endl;
        auto cur = p[i + 1] - p[i];
        while ((p[j + 1] - p[i]) * (cur)-eps > (p[j] - p[i]) * (cur))
        {
            j = (j % n) + 1;
        }
        while (mul(p[i + 1], p[k], p[i]) < mul(p[i + 1], p[k + 1], p[i]) - eps)
        {
            k = (k % n) + 1;
        }
        if (h == -1)
            h = k;
        while ((p[h + 1] - p[i]) * (cur) < (p[h] - p[i]) * (cur)-eps)
        {
            h = (h % n) + 1;
        }
        // cout << j << " " << k << " " << h << endl;
        double len = ((p[j] - p[i]) * (cur) - (p[h] - p[i]) * (cur)) / dis(p[i], p[i + 1]);
        double wid = distoline(p[i], p[i + 1], p[k]);
        // cout << len * wid << "\n";
        if (ret > len * wid)
        {
            ret = len * wid;
            ansi = array{i, j, k, h};
        }
    }
    matrix<T> res;
    {
        auto [i, j, k, h] = ansi;
        auto t = (p[i + 1] - p[i]) / dis(p[i + 1], p[i]);  // 矩形一条边的单位向量
        auto x = t * (t * (p[j] - p[i]));                  // 最右边的点的在该单位向量上的影射
        auto y = rot(t) * distoline(p[i], p[i + 1], p[k]); // 矩形的宽的向量
        auto z = t * (t * (p[h] - p[i]));                  // 最左边的点的在该单位向量上的影射

        res.a = (x + p[i]);
        res.b = (x + y + p[i]);
        res.c = (z + y + p[i]);
        res.d = (z + p[i]);
        res.area = ret;
    }
    return res; // 按照逆时针排序矩形四个点
}
void solve()
{
    int n;
    cin >> n;
    vector<point<ll>> a(n + 1);
    rep(i, 1, n)
    {
        cin >> a[i];
    }
    auto h = graham(a);
    vector<int> cnt(n + 1);
    sort(all(a), cmp<ll>);
    vector<point<ll>> b(1);
    int j = 1;
    for (int i = 1; i <= n; i++)
    {
        if (a[i] == h[j])
        {
            j++;
            continue;
        }
        cnt[i]++;
        // b.push_back(a[i]);
    }
    for (int i = n; i > 0; i--)
    {
        if (a[i] == h[j])
        {
            j++;
            continue;
        }
        cnt[i]++;
    }
    for (int i = 1; i <= n; i++)
    {
        if (cnt[i] == 2)
        {
            b.push_back(a[i]);
        }
    }
    if (b.size() == 1)
    {
        cout << -1 << "\n";
        return;
    }
    assert(b.size() - 1 == n - h.size() + 2);
    auto h2 = graham(b, 0);
    j = 1;
    n = h.size() - 2;
    int m = h2.size() - 1;
    // cout << n << m << endl;
    auto nxt = [&](int x)
    {
        return x % n + 1;
    };
    auto ret = area(h);
    ll mn = 2e18;
    if (m == 1)
    {
        for (int i = 1; i <= n; i++)
        {
            mn = min(mn, mul(h[i], h[nxt(i)], h2[1]));
        }
    }
    else
    {
        for (int i = 1; i <= n; i++)
        {
            while ((mul(h[i], h[nxt(i)], h2[j])) >= mul(h[i], h[nxt(i)], h2[j % m + 1]))
            {
                j = j % m + 1;
            }
            mn = min(mn, mul(h[i], h[nxt(i)], h2[j]));
        }
    }
    // for (int i = 1; i < h2.size(); i++)
    // {
    //     while (mul(h[j], h[nxt(j)], h2[i]) >= mul(h[nxt(j)], h[nxt(nxt(j))], h2[i]))
    //     {
    //         j = nxt(j);
    //     }
    //     mn = min(mn, mul(h[j], h[nxt(j)], h2[i]));
    // }
    // assert(mn >= 0);
    cout << ret - mn << endl;
}
int main()
{
    iofast;
    int t = 1;
    cin >> t;
    while (t--)
        solve();
    // cout << endl;
    //  system("pause");
}

Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 100
Accepted
time: 0ms
memory: 3616kb

input:

2
6
-2 0
1 -2
5 2
0 4
1 2
3 1
4
0 0
1 0
0 1
1 1

output:

40
-1

result:

ok 2 lines

Test #2:

score: 0
Accepted
time: 1ms
memory: 3648kb

input:

10
243
-494423502 -591557038
-493438474 -648991734
-493289308 -656152126
-491185085 -661710614
-489063449 -666925265
-464265894 -709944049
-447472922 -737242534
-415977509 -773788538
-394263365 -797285016
-382728841 -807396819
-373481975 -814685302
-368242265 -818267002
-344482838 -833805545
-279398...

output:

2178418010787347715
1826413114144932145
1651687576234220014
1883871859778998985
2119126281997959892
894016174881844630
2271191316922158910
1998643358049669416
1740474221286618711
1168195646932543192

result:

ok 10 lines

Test #3:

score: 0
Accepted
time: 27ms
memory: 3968kb

input:

1000
125
64661186 -13143076
302828013 -185438065
-418713797 -191594241
430218126 -397441626
354327250 -836704374
149668812 -598584998
311305970 66790541
199720625 -592356787
468137 -584752683
258775829 96211747
-358669612 -134890109
-129221188 -998432368
-277309896 -140056561
356901185 420557649
-51...

output:

1986320445246155278
1900093336073022078
1612088392301142476
2012259136539173407
1819942017252118749
1772230185841892196
1164835025329039520
1527446155241140517
1807368432185303666
1236918659444944569
1306839249967484778
1984123720246784099
1868728080720036006
667458140583450322
2127932992585026491
4...

result:

ok 1000 lines

Test #4:

score: 0
Accepted
time: 32ms
memory: 3872kb

input:

10000
9
484630042 51929469
-40468396 -517784096
98214104 -103353239
629244333 -475172587
106398764 153884485
49211709 -44865749
1 10
166321833 -247717657
406208245 668933360
13
548702216 -631976459
37150086 -292461024
707804811 -486185860
239775286 -903166050
10096571 -541890068
686103484 558731937
...

output:

950319193795831919
1661025342421008544
1285164852091455548
1159924751776806668
1206071151805176722
794021230296144371
699991678992587791
1133990718508584290
1486311831172661605
984875884297072200
1327767982175057345
758247019006396699
1355381234262206155
1139262078529131471
1613462877860621700
12392...

result:

ok 10000 lines

Test #5:

score: 0
Accepted
time: 63ms
memory: 3808kb

input:

100
439
471536154 -312612104
155692036 -937312180
-461624056 -357636609
236656684 -911414873
-288656914 -74788431
-465779694 -381475149
-334197401 -903065737
491513067 -447615916
337664889 -852236281
-281689379 -53519178
-159101704 -920779200
-326159514 -95396204
21868593 -994282736
488425383 -41046...

output:

1973162724053130487
2054612790507830954
1726805687754843724
1699420177872986528
2129388571309147631
2198295137903288810
1697185883164440272
1219697450095721478
2027023581922285255
1674691247127206655
1673105966817209954
2179188692918747442
2146544318743443141
2230356305133660648
1676850321902993764
...

result:

ok 100 lines

Test #6:

score: 0
Accepted
time: 43ms
memory: 3708kb

input:

100
1362
-467257672 -466669
-467054869 -478108
-466973270 -481776
-466556983 -499770
-466329414 -508693
-466248017 -511805
-466158865 -513786
-466101273 -515035
-465927700 -518748
-465717624 -522106
-465303448 -528127
-465124548 -530726
-464649746 -536693
-464554872 -537799
-464478196 -538680
-46416...

output:

1666097696993497
1791366071767866
1806187278469532
1683419854733713
1685891971828916
1730190225081651
1787048201197565
1850308098208660
1710694884375502
1826363113637639
1816375352374659
2047431269497691
1549806516003854
1829438662895747
1678707854135065
1687423392883819
2121960009997761
16687219538...

result:

ok 100 lines

Test #7:

score: 0
Accepted
time: 29ms
memory: 6400kb

input:

2
62666
-486101704 -505730259
-486101698 -506082699
-486101689 -506111362
-486101682 -506126031
-486101528 -506293759
-486101259 -506556385
-486101196 -506613483
-486101154 -506648604
-486100935 -506831392
-486100631 -507083675
-486100470 -507199151
-486100233 -507368923
-486100193 -507397039
-48609...

output:

2178736946152219010
1825181940245096152

result:

ok 2 lines

Test #8:

score: 0
Accepted
time: 71ms
memory: 9384kb

input:

2
100000
301945097 76373292
467957663 -286424714
8245445 -597212507
-474204621 -708828667
184159460 105942538
443435905 -429212625
490658771 -382198656
82512047 -612522436
-228221388 -965890088
394789011 -145801151
-106120174 -528202395
428939626 -194437311
497429477 -527407728
365739746 -114818962
...

output:

2502889432701099511
2267250485735988121

result:

ok 2 lines

Test #9:

score: 0
Accepted
time: 70ms
memory: 8712kb

input:

2
100000
221128057 -975244780
-618765360 -785575858
422567455 -906331476
-988680318 -150037424
-929870145 367887908
-757813541 -652471177
291995621 -956419655
-785381507 619012026
468864522 -883270094
-588416522 808557973
859345881 511394814
988105866 153775152
216931298 -976186873
467050734 8842305...

output:

6283183114882825575
6283183188903854361

result:

ok 2 lines

Test #10:

score: 0
Accepted
time: 0ms
memory: 3560kb

input:

7
5
-1000000000 -1000000000
1000000000 -1000000000
1000000000 1000000000
1 0
-1 0
5
1000000000 1000000000
-1000000000 -1000000000
-2 0
-1 0
1 -1
6
1000000000 1000000000
-1000000000 -1000000000
-3 0
-1 0
0 -1
1 -1
4
-1000000000 -1000000000
1000000000 -1000000000
1000000000 1000000000
-1000000000 1000...

output:

4000000000000000000
7000000000
9000000001
-1
6000000002000000000
7999999998000000000
-1

result:

ok 7 lines

Extra Test:

score: -3
Extra Test Failed : Time Limit Exceeded on 2

input:

1
7
-12 -7
10 -3
-15 -12
-12 -15
-2 12
-10 -9
8 3

output:


result: