bzoj 4445 小凸想跑步

题目传送门

　　vjudge的快速通道

　　bzoj的快速通道

题目大意

　　问在一个凸多边形内找一个点，连接这个点和所有顶点，使得与0号顶点，1号顶点构成的三角形是最小的概率。

　　假设点的位置是$(x, y)$，那么可以用叉积来计算三角形的面积。

　　这样可以列出$n - 1$个不等式。

　　将每个化成形如$ax + by + c leqslant 0$的形式。

　　然后分类讨论（$b = 0的时候需要特殊处理$）将它转换成二维平面上的半平面。

　　接着做半平面交，算面积就好了。

Code

/**
 * bzoj
 * Problem#4445
 * Accepted
 * Time: 288ms
 * Memory: 20068k
 */
#include <algorithm>
#include <iostream>
#include <cassert>
#include <cstring>
#include <cstdio>
#include <cmath>
using namespace std;
typedef bool boolean;

const double eps = 1e-8;
#define check(_s) cerr << _s << endl;

inline int dcmp(double x) {
    if (fabs(x) <= eps)    return 0;
    return (x < 0) ? (-1) : (1);
}

typedef class Vector {
    public:
        double x, y;
        
        Vector(double x = 0.0, double y = 0.0):x(x), y(y) {    }
}Point, Vector;

ostream& operator << (ostream& os, Point a) {
    os << "(" << a.x << ", " << a.y << ")";
    return os;
}

Vector operator + (Vector a, Vector b) {
    return Vector(a.x + b.x, a.y + b.y);
}

Vector operator - (Vector a, Vector b) {
    return Vector(a.x - b.x, a.y - b.y);
}

Vector operator * (double x, Vector a) {
    return Vector(a.x * x, a.y * x);
}

double cross(Vector a, Vector b) {
    return a.x * b.y - a.y * b.x;
}

double dot(Vector a, Vector b) {
    return a.x * b.x + a.y * b.y;
}

Point getLineIntersection(Point A, Vector v, Point B, Vector u) {
    double t = (cross(B - A, u) / cross(v, u));
    return A + t * v;
}

typedef class Line {
    public:
        Point p;
        Vector v;
        double ang;
    
        Line() {    }
        Line(Point p, Vector v):p(p), v(v) {
            ang = atan2(v.y, v.x);
        }

        boolean operator < (Line b) const {
//            return dcmp(cross(v, b.v)) > 0;
            return ang < b.ang;
        }
}Line;

const int N = 1e5 + 5;

int n;
int st = 1, ed = 0, top;
Line *ls, *qs;
Point *ps;

inline void init() {
    scanf("%d", &n);
    ps = new Point[(n << 1) + 5];
    for (int i = 0; i < n; i++)
        scanf("%lf%lf", &ps[i].x, &ps[i].y);
}

inline void build() {
    int d;
    Point cur, nxt, vec;
    ls = new Line[(n << 1) + 5];
    vec = ps[1] - ps[0];
    ls[++top] = Line(ps[0], vec);
    double bx = vec.y, by = vec.x, bc = cross(vec, ps[1]), nx, ny, nc;
    for (int i = 1; i < n; i++) {
        cur = ps[i], nxt = ps[(i + 1) % n], vec = nxt - cur;
        nx = bx - vec.y, ny = by - vec.x, nc = bc - cross(vec, cur);
        d = dcmp(ny);
        if (!d) {
            d = dcmp(nx);
            if (!d)    continue;
            ls[++top] = Line(Point(-nc / nx, 0), Vector(0, -d));
            
        } else {
            nx /= ny, nc /= ny;
            ls[++top] = Line(Point(0, nc), Vector(-d, -nx * d));
        }
        ls[++top] = Line(cur, vec);
    }
}

boolean isCrossingOut(Line b, Line cur, Line a) {
    assert (dcmp(cross(cur.v, a.v)));
//    if (!dcmp(cross(cur.v, a.v)))    return true;
    Point p = getLineIntersection(cur.p, cur.v, a.p, a.v);
    return dcmp(cross(p - b.p, b.v)) > 0;
}

inline void HalfPlaneIntersection(int n) {
    sort(ls + 1, ls + n + 1);
    qs = new Line[n + 5];
    for (int i = 1; i <= n; i++) {
        while (st < ed && isCrossingOut(ls[i], qs[ed], qs[ed - 1]))    ed--;
        while (st < ed && isCrossingOut(ls[i], qs[st], qs[st + 1]))    st++;
        qs[++ed] = ls[i];
        if (st < ed && !dcmp(cross(qs[ed].v, qs[ed - 1].v))) {
            ed--;
            if (dcmp(cross(qs[ed].p - ls[i].v, ls[i].v)) < 0)
                qs[ed] = ls[i];
        }
    }
    while (st < ed && isCrossingOut(qs[st], qs[ed], qs[ed - 1])) ed--;
}

inline double PolygonArea(Point *ps, int n) {
    double rt = cross(ps[n], ps[1]);
    for (int i = 1; i < n; i++)
        rt += cross(ps[i], ps[i + 1]);
    return fabs(rt) / 2;
}

inline void solve() {
    double all = PolygonArea(ps - 1, n);
    build();
    HalfPlaneIntersection(top);
//    assert(st < ed - 1);
    if (st >= ed - 1) {
        puts("0.0000");
        return;
    }
    for (int i = st; i < ed; i++)
        ps[i] = getLineIntersection(qs[i].p, qs[i].v, qs[i + 1].p, qs[i + 1].v);
    ps[ed] = getLineIntersection(qs[ed].p, qs[ed].v, qs[st].p, qs[st].v);
    printf("%.4f", PolygonArea(ps + (st - 1), ed - st + 1) / all);
}

int main() {
    init();
    solve();
    return 0;
}