牛客网暑期ACM多校训练营（第三场）J 多边形与圆相交的面积

链接：https://www.nowcoder.com/acm/contest/141/J

题目描述

Eddy has graduated from college. Currently, he is finding his future job and a place to live. Since Eddy is currently living in Tien-long country, he wants to choose a place inside Tien-long country to live. Surprisingly, Tien-long country can be represented as a simple polygon on 2D-plane. More surprisingly, Eddy can choose any place inside Tien-long country to live. The most important thing Eddy concerns is the distance from his place to the working place. He wants to live neither too close nor too far to the working place. The more specific definition of "close" and "far" is related to working place

Eddy has M choices to work in the future. For each working place, it can be represented as a point on 2D-plane. And, for each working place, Eddy has two magic parameters P and Q such that if Eddy is going to work in this place, he will choose a place to live which is closer to the working place than $\frac{P}{Q}$ portion of all possible living place choices.

Now, Eddy is wondering that for each working place, how far will he lives to the working place. Since Eddy is now busy on deciding where to work on, you come to help him calculate the answers.

For example, if the coordinates of points of Tien-long country is (0,0), (2,0), (2, 2), (0, 2) in counter-clockwise order. And, one possible working place is at (1,1) and P=1, Q=2. Then, Eddy should choose a place to live which is closer to (1, 1) than half of the choices. The distance from the place Eddy will live to the working place will be about 0.7978845608.

输入描述:

The first line contains one positive integer N indicating the number of points of the polygon representing Tien-long country.
Each of following N lines contains two space-separated integer (xi,yi)

indicating the coordinate of i-th points. These points is given in clockwise or counter-clockwise order and form the polygon.

Following line contains one positive integer M indicating the number of possible working place Eddy can choose from.
Each of following M lines contains four space-separated integer xj,yj,P,Q, where (xj,yj) indicating the j-th working place is at (xj,yj) and

magic parameters is P and Q.
3<=N<=200
1<=M<=200
1<=P<Q<=200
|xi||yi||xj||yj|<=1000

It's guaranteed that the given points form a simple polygon.

输出描述:

Output M lines. For i-th line, output one number indicating the distance from the place Eddy will live to the i-th working place.

Absolutely or relatively error within 10^-6 will be considered correct.

输入例子:

输出例子:

0.797884560809

-->

示例1

输入

输出

0.797884560809

示例2

输入

输出

1.040111537176

0.868735603376

题意一个国家由n个点组成 m次询问每次给出一个工作地点(xj,yj) 从国家里选取居住地点要满足选取的点比国家内Q/P的点离工作地点更近问居住点到工作地点的距离

解析我们可以二分答案mid 然后判断以(xj,yj)为圆心mid为半径的圆与国家相交的面积与国家面积的比值二分下去。

AC代码偷得模板。。。。

 #include <bits/stdc++.h>

 #define LL long long

 #define PI 3.1415926535897932384626

 #define maxn 1000

 #define EXIT exit(0);

 #define DEBUG puts("Here is a BUG");

 #define CLEAR(name, init) memset(name, init, sizeof(name))

 const double eps = 1e-;

 const int MAXN = (int)1e9 + ;

 using namespace std;

 #define Vector Point

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

 struct Point {

     double x, y;

     Point(const Point& rhs): x(rhs.x), y(rhs.y) { } //拷贝构造函数

     Point(double x = 0.0, double y = 0.0): x(x), y(y) { }   //构造函数

     friend istream& operator >> (istream& in, Point& P) { return in >> P.x >> P.y; }

     friend ostream& operator << (ostream& out, const Point& P) { return out << P.x << ' ' << P.y; }

     friend Vector operator + (const Vector& A, const Vector& B) { return Vector(A.x+B.x, A.y+B.y); }

     friend Vector operator - (const Point& A, const Point& B) { return Vector(A.x-B.x, A.y-B.y); }

     friend Vector operator * (const Vector& A, const double& p) { return Vector(A.x*p, A.y*p); }

     friend Vector operator / (const Vector& A, const double& p) { return Vector(A.x/p, A.y/p); }

     friend bool operator == (const Point& A, const Point& B) { return dcmp(A.x-B.x) ==  && dcmp(A.y-B.y) == ; }

     friend bool operator < (const Point& A, const Point& B) { return A.x < B.x || (A.x == B.x && A.y < B.y); }

     void in(void) { scanf("%lf%lf", &x, &y); }

     void out(void) { printf("%lf %lf", x, y); }

 };

 template <class T> T sqr(T x) { return x * x;}

 double Dot(const Vector& A, const Vector& B) { return A.x*B.x + A.y*B.y; }  //点积

 double Length(const Vector& A){ return sqrt(Dot(A, A)); }

 double Angle(const Vector& A, const Vector& B) { return acos(Dot(A, B)/Length(A)/Length(B)); }  //向量夹角

 double Cross(const Vector& A, const Vector& B) { return A.x*B.y - A.y*B.x; }    //叉积

 double Area(const Point& A, const Point& B, const Point& C) { return fabs(Cross(B-A, C-A)); }

 Vector normal(Vector x) { return Point(-x.y, x.x) / Length(x);}

 double angle(Vector x) { return atan2(x.y, x.x);}

 Vector vecunit(Vector x){ return x / Length(x);} //单位向量

 struct Circle {

     Point c;    //圆心

     double r;   //半径

     Circle() { }

     Circle(const Circle& rhs): c(rhs.c), r(rhs.r) { }

     Circle(const Point& c, const double& r): c(c), r(r) { }

     Point point(double ang) const { return Point(c.x + cos(ang)*r, c.y + sin(ang)*r); } //圆心角所对应的点

     double area(void) const { return PI * r * r; }

 };

 struct Line {

     Point P;    //直线上一点

     Vector dir; //方向向量(半平面交中该向量左侧表示相应的半平面)

     double ang; //极角，即从x正半轴旋转到向量dir所需要的角（弧度）

     Line() { }  //构造函数

     Line(const Line& L): P(L.P), dir(L.dir), ang(L.ang) { }

     Line(const Point& P, const Vector& dir): P(P), dir(dir) { ang = atan2(dir.y, dir.x); }

     bool operator < (const Line& L) const { //极角排序

         return ang < L.ang;

     }

     Point point(double t) { return P + dir*t; }

 };

 bool InCircle(Point x, Circle c) { return dcmp(c.r*c.r - Length(c.c - x)*Length(c.c - x)) >= ;}

 Point GetIntersection(Line a, Line b) //线段交点

 {

     Vector u = a.P-b.P;

     double t = Cross(b.dir, u) / Cross(a.dir, b.dir);

     return a.P + a.dir*t;

 }

 bool OnSegment(Point p, Point a1, Point a2)

 {

     return dcmp(Cross(a1-p, a2-p)) ==  && dcmp(Dot(a1-p, a2-p)) < ;

 }

 int getSegCircleIntersection(Line L, Circle C, Point* sol)

 {

     Vector nor = normal(L.dir);

     Line pl = Line(C.c, nor);

     Point ip = GetIntersection(pl, L);

     double dis = Length(ip - C.c);

     if (dcmp(dis - C.r) > ) return ;

     Point dxy = vecunit(L.dir) * sqrt(C.r*C.r - dis*dis);

     int ret = ;

     sol[ret] = ip + dxy;

     if (OnSegment(sol[ret], L.P, L.point())) ret++;

     sol[ret] = ip - dxy;

     if (OnSegment(sol[ret], L.P, L.point())) ret++;

     return ret;

 }

 double SegCircleArea(Circle C, Point a, Point b) //线段切割圆

 {

     double a1 = angle(a - C.c);

     double a2 = angle(b - C.c);

     double da = fabs(a1 - a2);

     if (da > PI) da = PI * 2.0 - da;

     return dcmp(Cross(b - C.c, a - C.c)) * da * sqr(C.r) / 2.0;

 }

 double PolyCiclrArea(Circle C, Point *p, int n)//多边形与圆相交面积

 {

     double ret = 0.0;

     Point sol[];

     p[n] = p[];

     for(int i=;i<n;i++)

     {

         double t1, t2;

         int cnt = getSegCircleIntersection(Line(p[i], p[i+]-p[i]), C, sol);

         if (cnt == )

         {

             if (!InCircle(p[i], C) || !InCircle(p[i+], C)) ret += SegCircleArea(C, p[i], p[i+]);

             else ret += Cross(p[i+] - C.c, p[i] - C.c) / 2.0;

         }

         if (cnt == )

         {

             if (InCircle(p[i], C) && !InCircle(p[i+], C)) ret += Cross(sol[] - C.c, p[i] - C.c) / 2.0, ret += SegCircleArea(C, sol[], p[i+]);

             else ret += SegCircleArea(C, p[i], sol[]), ret += Cross(p[i+] - C.c, sol[] - C.c) / 2.0;

         }

         if (cnt == )

         {

             if ((p[i] < p[i + ]) ^ (sol[] < sol[])) swap(sol[], sol[]);

             ret += SegCircleArea(C, p[i], sol[]);

             ret += Cross(sol[] - C.c, sol[] - C.c) / 2.0;

             ret += SegCircleArea(C, sol[], p[i+]);

         }

     }

     return fabs(ret);

 }

 double PolygonArea(Point *po, int n) {

     double area = 0.0;

     for(int i = ; i < n-; i++) {

         area += Cross(po[i]-po[], po[i+]-po[]);

     }

     return area * 0.5;

 }

 Point a[], b;

 double p, q;

 int main(){

     int n, m;

     scanf("%d", &n);

     for(int i=;i<n;i++) a[i].in();

     double ar = fabs(PolygonArea(a, n));

     scanf("%d", &m);

     for(int i=;i<m;i++){

         b.in();

         scanf("%lf%lf", &p, &q);

         double l = , r = , num = , mid, aa = -p/q;

         while(num--){

             mid = (l+r)/;

             Circle yuan(b, mid);

             if(PolyCiclrArea(yuan, a, n)/ar < aa) l = mid;

             else r = mid;

         }

         printf("%.10f\n", mid);

     }

     return ;

 }