bzoj 4445 小凸想跑步 - 半平面交

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-;
 #define check(_s) cerr << _s << endl;

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

 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 + ;

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

 inline void init() {
     scanf("%d", &n);
     ps = new Point[(n << ) + ];
     for (int i = ; 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 << ) + ];
     vec = ps[] - ps[];
     ls[++top] = Line(ps[], vec);
     double bx = vec.y, by = vec.x, bc = cross(vec, ps[]), nx, ny, nc;
     for (int i = ; i < n; i++) {
         cur = ps[i], nxt = ps[(i + ) % 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, ), Vector(, -d));

         } else {
             nx /= ny, nc /= ny;
             ls[++top] = Line(Point(, 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)) > ;
 }

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

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

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

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