[ICPC2020南京I] Interested in Skiing - dp,计算几何
Description
二维平面上有若干条线段和两条直线 x=-m, x=m 作为障碍，从 \((0,-\infty)\) 走到 \((0,\infty)\)，不能碰到任何障碍。\(y\) 方向速度恒定为 \(+v_y\)，\(x\) 方向速度可以随意调节，但其绝对值有一个上界 \(a\)。求最小的 \(a\) 是多少。
Solution
最优策略一定是经过若干条线段的端点，因此拿着所有端点做个 dp 即可，过程中需要用到判交，暴力判断即可，总体时间复杂度 \(O(n^3)\)
#include <bits/stdc++.h>
using namespace std;
namespace geo
{
#define mp make_pair
#define fi first
#define se second
#define pb push_back
typedef double db;
const db eps = 1e-6;
const db pi = acos(-1);
int sign(db k)
{
if (k > eps)
return 1;
else if (k < -eps)
return -1;
return 0;
}
int cmp(db p1, db p2) { return sign(p1 - p2); }
int inmid(db p1, db p2, db p3) { return sign(p1 - p3) * sign(p2 - p3) <= 0; }
struct point
{
db x, y;
point operator+(const point &p1) const { return (point){p1.x + x, p1.y + y}; }
point operator-(const point &p1) const { return (point){x - p1.x, y - p1.y}; }
point operator*(db p1) const { return (point){x * p1, y * p1}; }
point operator/(db p1) const { return (point){x / p1, y / p1}; }
int operator==(const point &p1) const { return cmp(x, p1.x) == 0 && cmp(y, p1.y) == 0; }
// 逆时针旋转
point turn(db p1) { return (point){x * cos(p1) - y * sin(p1), x * sin(p1) + y * cos(p1)}; }
point turn90() { return (point){-y, x}; }
bool operator<(const point p1) const
{
int a = cmp(x, p1.x);
if (a == -1)
return 1;
else if (a == 1)
return 0;
else
return cmp(y, p1.y) == -1;
}
db abs() { return sqrt(x * x + y * y); }
db abs2() { return x * x + y * y; }
db dis(point p1) { return ((*this) - p1).abs(); }
point unit()
{
db w = abs();
return (point){x / w, y / w};
}
void scan()
{
double p1, p2;
cin >> p1 >> p2;
x = p1;
y = p2;
}
void print() { printf("%.11lf %.11lf\n", x, y); }
db getw() { return atan2(y, x); }
point getdel()
{
if (sign(x) == -1 || (sign(x) == 0 && sign(y) == -1))
return (*this) * (-1);
else
return (*this);
}
int getP() const { return sign(y) == 1 || (sign(y) == 0 && sign(x) == -1); }
};
int inmid(point p1, point p2, point p3) { return inmid(p1.x, p2.x, p3.x) && inmid(p1.y, p2.y, p3.y); }
db cross(point p1, point p2) { return p1.x * p2.y - p1.y * p2.x; }
db dot(point p1, point p2) { return p1.x * p2.x + p1.y * p2.y; }
db rad(point p1, point p2) { return atan2(cross(p1, p2), dot(p1, p2)); }
// -pi -> pi
int compareangle(point p1, point p2)
{
return p1.getP() < p2.getP() || (p1.getP() == p2.getP() && sign(cross(p1, p2)) > 0);
}
point proj(point p1, point p2, point q)
{ // q 到直线 p1,p2 的投影
point k = p2 - p1;
return p1 + k * (dot(q - p1, k) / k.abs2());
}
point reflect(point p1, point p2, point q) { return proj(p1, p2, q) * 2 - q; }
int clockwise(point p1, point p2, point p3)
{ // p1 p2 p3 逆时针 1 顺时针 -1 否则 0
return sign(cross(p2 - p1, p3 - p1));
}
int checkLL(point p1, point p2, point p3, point p4)
{ // 求直线 (L) 线段 (S)p1,p2 和 p3,p4 的交点
return cmp(cross(p3 - p1, p4 - p1), cross(p3 - p2, p4 - p2)) != 0;
}
point getLL(point p1, point p2, point p3, point p4)
{
db w1 = cross(p1 - p3, p4 - p3), w2 = cross(p4 - p3, p2 - p3);
return (p1 * w2 + p2 * w1) / (w1 + w2);
}
int intersect(db l1, db r1, db l2, db r2)
{
if (l1 > r1)
swap(l1, r1);
if (l2 > r2)
swap(l2, r2);
return cmp(r1, l2) == 1 && cmp(r2, l1) == 1;
}
int checkSS(point p1, point p2, point p3, point p4)
{
return intersect(p1.x, p2.x, p3.x, p4.x) && intersect(p1.y, p2.y, p3.y, p4.y) &&
sign(cross(p3 - p1, p4 - p1)) * sign(cross(p3 - p2, p4 - p2)) < 0 &&
sign(cross(p1 - p3, p2 - p3)) * sign(cross(p1 - p4, p2 - p4)) < 0;
}
db disSP(point p1, point p2, point q)
{
point p3 = proj(p1, p2, q);
if (inmid(p1, p2, p3))
return q.dis(p3);
else
return min(q.dis(p1), q.dis(p2));
}
db disSS(point p1, point p2, point p3, point p4)
{
if (checkSS(p1, p2, p3, p4))
return 0;
else
return min(min(disSP(p1, p2, p3), disSP(p1, p2, p4)), min(disSP(p3, p4, p1), disSP(p3, p4, p2)));
}
int onS(point p1, point p2, point q) { return inmid(p1, p2, q) && sign(cross(p1 - q, p2 - p1)) == 0; }
struct line
{
// p[0]->p[1]
point p[2];
line(point p1 = {0, 0}, point p2 = {0, 0})
{
p[0] = p1;
p[1] = p2;
}
point &operator[](int k) { return p[k]; }
int include(point k) { return sign(cross(p[1] - p[0], k - p[0])) > 0; }
point dir() { return p[1] - p[0]; }
line push()
{ // 向外 ( 左手边 ) 平移 eps
const db eps = 1e-6;
point delta = (p[1] - p[0]).turn90().unit() * eps;
return {p[0] - delta, p[1] - delta};
}
};
point getLL(line p1, line p2) { return getLL(p1[0], p1[1], p2[0], p2[1]); }
int parallel(line p1, line p2) { return sign(cross(p1.dir(), p2.dir())) == 0; }
int sameDir(line p1, line p2) { return parallel(p1, p2) && sign(dot(p1.dir(), p2.dir())) == 1; }
int operator<(line p1, line p2)
{
if (sameDir(p1, p2))
return p2.include(p1[0]);
return compareangle(p1.dir(), p2.dir());
}
int checkpos(line p1, line p2, line p3) { return p3.include(getLL(p1, p2)); }
}
typedef geo::point vec2;
typedef geo::line line;
using geo::checkSS;
bool cmp(const vec2 &lhs, const vec2 &rhs)
{
return lhs.y < rhs.y;
}
signed main()
{
ios::sync_with_stdio(false);
int n, m;
double vy;
cin >> n >> m >> vy;
vector<line> l(n + 2);
for (int i = 1; i <= n; i++)
l[i].p[0].scan(), l[i].p[1].scan();
vector<vec2> p(2 * n + 2);
p[0] = {0, -1e100};
for (int i = 1; i <= n; i++)
p[2 * i - 1] = l[i].p[0], p[2 * i] = l[i].p[1];
p[2 * n + 1] = {0, 1e100};
sort(p.begin(), p.end(), cmp);
vector<double> f(2 * n + 2);
for (int i = 1; i <= 2 * n + 1; i++)
{
f[i] = 1e100;
if (abs(p[i].x) >= m)
continue;
for (int j = 0; j < i; j++)
{
if (p[i].y == p[j].y)
continue;
int flag = 1;
for (int k = 1; k <= n; k++)
{
if (checkSS(l[k].p[0], l[k].p[1], p[j], p[i]))
{
flag = 0;
break;
}
}
if (flag)
{
f[i] = min(f[i], max(f[j], abs(p[i].x - p[j].x) / (p[i].y - p[j].y + 1e-18)));
}
}
}
if (f[2 * n + 1] <= 1e20)
cout << fixed << setprecision(16) << f[2 * n + 1] * vy << endl;
else
cout << -1 << endl;
}