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

  1 /**
  2  * bzoj
  3  * Problem#4445
  4  * Accepted
  5  * Time: 288ms
  6  * Memory: 20068k
  7  */
  8 #include <algorithm>
  9 #include <iostream>
 10 #include <cassert>
 11 #include <cstring>
 12 #include <cstdio>
 13 #include <cmath>
 14 using namespace std;
 15 typedef bool boolean;
 16 
 17 const double eps = 1e-8;
 18 #define check(_s) cerr << _s << endl;
 19 
 20 inline int dcmp(double x) {
 21     if (fabs(x) <= eps)    return 0;
 22     return (x < 0) ? (-1) : (1);
 23 }
 24 
 25 typedef class Vector {
 26     public:
 27         double x, y;
 28         
 29         Vector(double x = 0.0, double y = 0.0):x(x), y(y) {    }
 30 }Point, Vector;
 31 
 32 ostream& operator << (ostream& os, Point a) {
 33     os << "(" << a.x << ", " << a.y << ")";
 34     return os;
 35 }
 36 
 37 Vector operator + (Vector a, Vector b) {
 38     return Vector(a.x + b.x, a.y + b.y);
 39 }
 40 
 41 Vector operator - (Vector a, Vector b) {
 42     return Vector(a.x - b.x, a.y - b.y);
 43 }
 44 
 45 Vector operator * (double x, Vector a) {
 46     return Vector(a.x * x, a.y * x);
 47 }
 48 
 49 double cross(Vector a, Vector b) {
 50     return a.x * b.y - a.y * b.x;
 51 }
 52 
 53 double dot(Vector a, Vector b) {
 54     return a.x * b.x + a.y * b.y;
 55 }
 56 
 57 Point getLineIntersection(Point A, Vector v, Point B, Vector u) {
 58     double t = (cross(B - A, u) / cross(v, u));
 59     return A + t * v;
 60 }
 61 
 62 typedef class Line {
 63     public:
 64         Point p;
 65         Vector v;
 66         double ang;
 67     
 68         Line() {    }
 69         Line(Point p, Vector v):p(p), v(v) {
 70             ang = atan2(v.y, v.x);
 71         }
 72 
 73         boolean operator < (Line b) const {
 74 //            return dcmp(cross(v, b.v)) > 0;
 75             return ang < b.ang;
 76         }
 77 }Line;
 78 
 79 const int N = 1e5 + 5;
 80 
 81 int n;
 82 int st = 1, ed = 0, top;
 83 Line *ls, *qs;
 84 Point *ps;
 85 
 86 inline void init() {
 87     scanf("%d", &n);
 88     ps = new Point[(n << 1) + 5];
 89     for (int i = 0; i < n; i++)
 90         scanf("%lf%lf", &ps[i].x, &ps[i].y);
 91 }
 92 
 93 inline void build() {
 94     int d;
 95     Point cur, nxt, vec;
 96     ls = new Line[(n << 1) + 5];
 97     vec = ps[1] - ps[0];
 98     ls[++top] = Line(ps[0], vec);
 99     double bx = vec.y, by = vec.x, bc = cross(vec, ps[1]), nx, ny, nc;
100     for (int i = 1; i < n; i++) {
101         cur = ps[i], nxt = ps[(i + 1) % n], vec = nxt - cur;
102         nx = bx - vec.y, ny = by - vec.x, nc = bc - cross(vec, cur);
103         d = dcmp(ny);
104         if (!d) {
105             d = dcmp(nx);
106             if (!d)    continue;
107             ls[++top] = Line(Point(-nc / nx, 0), Vector(0, -d));
108             
109         } else {
110             nx /= ny, nc /= ny;
111             ls[++top] = Line(Point(0, nc), Vector(-d, -nx * d));
112         }
113         ls[++top] = Line(cur, vec);
114     }
115 }
116 
117 boolean isCrossingOut(Line b, Line cur, Line a) {
118     assert (dcmp(cross(cur.v, a.v)));
119 //    if (!dcmp(cross(cur.v, a.v)))    return true;
120     Point p = getLineIntersection(cur.p, cur.v, a.p, a.v);
121     return dcmp(cross(p - b.p, b.v)) > 0;
122 }
123 
124 inline void HalfPlaneIntersection(int n) {
125     sort(ls + 1, ls + n + 1);
126     qs = new Line[n + 5];
127     for (int i = 1; i <= n; i++) {
128         while (st < ed && isCrossingOut(ls[i], qs[ed], qs[ed - 1]))    ed--;
129         while (st < ed && isCrossingOut(ls[i], qs[st], qs[st + 1]))    st++;
130         qs[++ed] = ls[i];
131         if (st < ed && !dcmp(cross(qs[ed].v, qs[ed - 1].v))) {
132             ed--;
133             if (dcmp(cross(qs[ed].p - ls[i].v, ls[i].v)) < 0)
134                 qs[ed] = ls[i];
135         }
136     }
137     while (st < ed && isCrossingOut(qs[st], qs[ed], qs[ed - 1])) ed--;
138 }
139 
140 inline double PolygonArea(Point *ps, int n) {
141     double rt = cross(ps[n], ps[1]);
142     for (int i = 1; i < n; i++)
143         rt += cross(ps[i], ps[i + 1]);
144     return fabs(rt) / 2;
145 }
146 
147 inline void solve() {
148     double all = PolygonArea(ps - 1, n);
149     build();
150     HalfPlaneIntersection(top);
151 //    assert(st < ed - 1);
152     if (st >= ed - 1) {
153         puts("0.0000");
154         return;
155     }
156     for (int i = st; i < ed; i++)
157         ps[i] = getLineIntersection(qs[i].p, qs[i].v, qs[i + 1].p, qs[i + 1].v);
158     ps[ed] = getLineIntersection(qs[ed].p, qs[ed].v, qs[st].p, qs[st].v);
159     printf("%.4f", PolygonArea(ps + (st - 1), ed - st + 1) / all);
160 }
161 
162 int main() {
163     init();
164     solve();
165     return 0;
166 }
posted @ 2018-03-17 16:28  阿波罗2003  阅读(265)  评论(0)  编辑  收藏  举报