UVALive 4428 Solar Eclipse --計算幾何，圓相交

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題意：平面上有一些半徑為R的圓，現在要在滿足不與現有圓相交的條件下放入一個圓，求這個圓能放的位置的圓心到原點的最短距離。

解法：我們將半徑擴大一倍，R = 2*R，那麼在每個圓上或圓外的位置都可以放圓心了。

首先特判放到原點可不可以，如果不可以，再將所有圓的圓心與原點的直線與該圓相交的點放入隊列，再將所有圓兩兩相交的點放入隊列，然後處理整個隊列，一一判斷這些點行不行，可以證明，最優點一定在這些裡面。

如果有一個圓的圓心在(0,0)點，那麼要特判一下，因為此時圓心與原點連的直線長度為0，對於這種情況，我們判一下(R,0)這個就行了。

代碼：

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>#include <algorithm>#include <vector>#define Mod 1000000007#define eps 1e-7using 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 borderint GetCircleCircleIntersection(Circle C1, Circle C2, vector<Point>& sol)  //return 交點個數{    double d = Length(C1.c - C2.c);    if(dcmp(d) == 0){        if(dcmp(C1.r - C2.r) == 0) return -1;  //兩圓重合        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的極角    double da = acos((sqr(C1.r) + sqr(d) - sqr(C2.r)) / (2*C1.r*d)); //C1C2到C1P1的極角    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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UVALive 4428 Solar Eclipse --計算幾何，圓相交