QOJ.ac
QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#723700 | #9520. Concave Hull | nan01 | TL | 71ms | 9384kb | C++23 | 14.8kb | 2024-11-07 23:47:29 | 2024-11-07 23:47:29 |
Judging History
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