HDU 5130 Signal Interference（计算几何）

思路：把给出的公式化简...然后就会发现是个神奇的圆，然后就是多边形和圆的面积的并

#include<bits/stdc++.h>
#define eps 1e-8
using namespace std;
#define N 100017

struct Point{
    double x,y;
    Point(double x=0, double y=0):x(x),y(y) {}
    void input() { scanf("%lf%lf",&x,&y); }
};
typedef Point Vector;
struct Circle{
    Point c;
    double r;
    Circle(){}
    Circle(Point c,double r):c(c),r(r) {}
    Point point(double a) { return Point(c.x + cos(a)*r, c.y + sin(a)*r); }
    void input() { scanf("%lf%lf%lf",&c.x,&c.y,&r); }
};
int dcmp(double x) {
    if(x < -eps) return -1;
    if(x > eps) return 1;
    return 0;
}
template <class T> T sqr(T x) { return x * x;}
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 * (Vector A, double p) { return Vector(A.x*p, A.y*p); }
Vector operator / (Vector A, double p) { return Vector(A.x/p, A.y/p); }
bool operator < (const Point& a, const Point& b) { return a.x < b.x || (a.x == b.x && a.y < b.y); }
bool operator >= (const Point& a, const Point& b) { return a.x >= b.x && a.y >= b.y; }
bool operator <= (const Point& a, const Point& b) { return a.x <= b.x && a.y <= b.y; }
bool operator == (const Point& a, const Point& b) { return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0; }
double Dot(Vector A, Vector B) { return A.x*B.x + A.y*B.y; }
double Length(Vector A) { return sqrt(Dot(A, A)); }
double Angle(Vector A, Vector B) { return acos(Dot(A, B) / Length(A) / Length(B)); }
double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; }
Vector VectorUnit(Vector x){ return x / Length(x);}
Vector Normal(Vector x) { return Point(-x.y, x.x) / Length(x);}
double angle(Vector v) { return atan2(v.y, v.x); }

bool OnSegment(Point P, Point A, Point B) {
    return dcmp(Cross(A-P,B-P)) == 0 && dcmp(Dot(A-P,B-P)) < 0;
}
double DistanceToSeg(Point P, Point A, Point B)
{
    if(A == B) return Length(P-A);
    Vector v1 = B-A, v2 = P-A, v3 = P-B;
    if(dcmp(Dot(v1, v2)) < 0) return Length(v2);
    if(dcmp(Dot(v1, v3)) > 0) return Length(v3);
    return fabs(Cross(v1, v2)) / Length(v1);
}
double DistanceToLine(Point P, Point A, Point B){
    Vector v1 = B-A, v2 = P-A;
    return fabs(Cross(v1,v2)) / Length(v1);
}
Point DisP(Point A, Point B){
    return Length(B-A);
}
bool SegmentIntersection(Point A,Point B,Point C,Point D) {
    return max(A.x,B.x) >= min(C.x,D.x) &&
           max(C.x,D.x) >= min(A.x,B.x) &&
           max(A.y,B.y) >= min(C.y,D.y) &&
           max(C.y,D.y) >= min(A.y,B.y) &&
           dcmp(Cross(C-A,B-A)*Cross(D-A,B-A)) <= 0 &&
           dcmp(Cross(A-C,D-C)*Cross(B-C,D-C)) <= 0;
}

Point Zero = Point(0,0);
double TriAngleCircleInsection(Circle C, Point A, Point B)
{
    Vector OA = A-C.c, OB = B-C.c;
    Vector BA = A-B, BC = C.c-B;
    Vector AB = B-A, AC = C.c-A;
    double DOA = Length(OA), DOB = Length(OB),DAB = Length(AB), r = C.r;
    if(dcmp(Cross(OA,OB)) == 0) return 0;
    if(dcmp(DOA-C.r) < 0 && dcmp(DOB-C.r) < 0) return Cross(OA,OB)*0.5;
    else if(DOB < r && DOA >= r) {
        double x = (Dot(BA,BC) + sqrt(r*r*DAB*DAB-Cross(BA,BC)*Cross(BA,BC)))/DAB;
        double TS = Cross(OA,OB)*0.5;
        return asin(TS*(1-x/DAB)*2/r/DOA)*r*r*0.5+TS*x/DAB;
    }
    else if(DOB >= r && DOA < r) {
        double y = (Dot(AB,AC)+sqrt(r*r*DAB*DAB-Cross(AB,AC)*Cross(AB,AC)))/DAB;
        double TS = Cross(OA,OB)*0.5;
        return asin(TS*(1-y/DAB)*2/r/DOB)*r*r*0.5+TS*y/DAB;
    }
    else if(fabs(Cross(OA,OB)) >= r*DAB || Dot(AB,AC) <= 0 || Dot(BA,BC) <= 0) {
        if(Dot(OA,OB) < 0) {
            if(Cross(OA,OB) < 0) return (-acos(-1.0)-asin(Cross(OA,OB)/DOA/DOB))*r*r*0.5;
            else                 return ( acos(-1.0)-asin(Cross(OA,OB)/DOA/DOB))*r*r*0.5;
        }
        else                     return asin(Cross(OA,OB)/DOA/DOB)*r*r*0.5;
    }
    else {
        double x = (Dot(BA,BC)+sqrt(r*r*DAB*DAB-Cross(BA,BC)*Cross(BA,BC)))/DAB;
        double y = (Dot(AB,AC)+sqrt(r*r*DAB*DAB-Cross(AB,AC)*Cross(AB,AC)))/DAB;
        double TS = Cross(OA,OB)*0.5;
        return (asin(TS*(1-x/DAB)*2/r/DOA)+asin(TS*(1-y/DAB)*2/r/DOB))*r*r*0.5 + TS*((x+y)/DAB-1);
    }
}
//data segment
Point p[507],A,B;
//data ends

int main()
{
    int n,i,j,cs = 1;
    double k;
    while(scanf("%d%lf",&n,&k)!=EOF)
    {
        for(i=1;i<=n;i++) p[i].input();
        A.input(), B.input(), p[n+1] = p[1];
        double D = (2.0*k*k*A.x-2.0*B.x)/(1.0-k*k);
        double E = (2.0*k*k*A.y-2.0*B.y)/(1.0-k*k);
        double F = (B.x*B.x+B.y*B.y-k*k*(A.x*A.x+A.y*A.y))/(1.0-k*k);
        Circle C = Circle(Point(-D*0.5,-E*0.5),sqrt(D*D+E*E-4.0*F)*0.5);
        double ans = 0.0;
        for(i=1;i<=n;i++)
            ans += TriAngleCircleInsection(C, p[i], p[i+1]);
        printf("Case %d: %.10f\n",cs++,fabs(ans));
    }
    return 0;
}

Description

Two countries A-Land and B-Land are at war. The territory of A-Land is a simple polygon with no more than 500 vertices. For military use, A-Land constructed a radio tower (also written as A), and it's so powerful that the whole country was under its signal. To interfere A-Land's communication, B-Land decided to build another radio tower (also written as B). According to an accurate estimation, for any point P, if the euclidean distance between P and B is no more than k (0.2 ≤ k < 0.8) times of the distance between P and A, then point P is not able to receive clear signals from A, i.e. be interfered. Your task is to calculate the area in A-Land's territory that are under B-Land's interference.

 

Input

There are no more than 100 test cases in the input. 

In each test case, firstly you are given a positive integer N indicating the amount of vertices on A-Land's territory, and an above mentioned real number k, which is rounded to 4 digits after the decimal point. 

Then N lines follow. Each line contains two integers x and y (|x|, |y| ≤ 1000), indicating a vertex's coordinate on A's territory, in counterclockwise or clockwise order. 

The last two lines of a test case give radio tower A and B's coordinates in the same form as vertexes' coordinates. You can assume that A is not equal to B.

 

Output

For each test case, firstly output the case number, then output your answer in one line following the format shown in sample. Please note that there is a blank after the ':'. 

Your solution will be accepted if its absolute error or relative error is no more than 10-6. 

This problem is special judged.

 

Sample Input

        
        
        
        

         
         
         
         4 0.5000
-1 -1
1 -1
1 1
-1 1
0 0
-1 0 
        
        
        
        

 

Sample Output

        
        
        
        

         
         
         
         Case 1: 0.2729710441