UVALive 4428 Solar Eclipse --计算几何，圆相交

题意：平面上有一些半径为R的圆，现在要在满足不与现有圆相交的条件下放入一个圆，求这个圆能放的位置的圆心到原点的最短距离。

解法：我们将半径扩大一倍，R = 2*R，那么在每个圆上或圆外的位置都可以放圆心了。

首先特判放到原点可不可以，如果不可以，再将所有圆的圆心与原点的直线与该圆相交的点放入队列，再将所有圆两两相交的点放入队列，然后处理整个队列，一一判断这些点行不行，可以证明，最优点一定在这些里面。

如果有一个圆的圆心在(0,0)点，那么要特判一下，因为此时圆心与原点连的直线长度为0，对于这种情况，我们判一下(R,0)这个就行了。

代码：

#include <iostream>

#include <cstdio>

#include <cstring>

#include <cstdlib>

#include <cmath>

#include <algorithm>

#include <vector>

#define Mod 1000000007

#define eps 1e-7

using namespace std;



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; }

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



bool InCircle(Point x, Circle c) { return dcmp(c.r - Length(c.c-x)) > 0; } //not in border

int GetCircleCircleIntersection(Circle C1, Circle C2, vector<Point>& sol)  //return 交点个数

{

    double d = Length(C1.c - C2.c);

    if(dcmp(d) == 0){

        if(dcmp(C1.r - C2.r) == 0) return -1;  //两圆重合

        return 0;

    }

    if(dcmp(C1.r + C2.r - d) < 0) return 0;

    if(dcmp(fabs(C1.r - C2.r) - d) > 0) return 0;



    double a = angle(C2.c - C1.c);             //向量C1C2的极角

    double da = acos((sqr(C1.r) + sqr(d) - sqr(C2.r)) / (2*C1.r*d)); //C1C2到C1P1的极角



    Point p1 = C1.point(a-da), p2 = C1.point(a+da);

    sol.push_back(p1);

    if(p1 == p2) return 1;

    sol.push_back(p2);

    return 2;

}

double DISP(Point p) { return sqrt(p.x*p.x+p.y*p.y); }



Circle C[106],sC[106];

int n;



bool check(Point now) {

    for(int i=1;i<=n;i++) {

        if(InCircle(now,C[i]))

            return false;

    }

    return true;

}





int main()

{

    int i,j;

    double R;

    while(scanf("%d%lf",&n,&R)!=EOF && n+R)

    {

        for(i=1;i<=n;i++)

        {

            scanf("%lf%lf",&C[i].c.x,&C[i].c.y), C[i].r = 2.0*R;

            sC[i] = C[i], sC[i].r = R;

        }

        vector<Point> sec;

        sec.clear();

        for(i=1;i<=n;i++) {

            for(j=i+1;j<=n;j++)

                GetCircleCircleIntersection(C[i],C[j],sec);

        }

        double Mini = Mod;

        if(check(Point(0,0))) { printf("%.6f\n",0.0); continue; }

        for(i=1;i<=n;i++) {

            if(dcmp(DISP(C[i].c)) == 0) {

                if(check(Point(2*R,0))) Mini = min(Mini,2*R);

                continue;

            }

            sec.push_back(Point(C[i].c+C[i].c*(-2.0*R/DISP(C[i].c))));

            sec.push_back(Point(C[i].c+C[i].c*(2.0*R/DISP(C[i].c))));

        }

        for(i=0;i<sec.size();i++)

            if(check(sec[i])) Mini = min(Mini,DISP(sec[i]));

        printf("%.6f\n",Mini);

    }

    return 0;

}

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