HDU 3561 How many times 【计算几何】

判断任两圆的交点和每个圆的圆心即可

#include<iostream>#include<vector>#include<math.h>using namespace std;#define eps 1e-8const int maxn=105;struct point{    double x,y;};struct circle{    point p;    double r;}c[maxn];vector<point>v;double dis(point p1,point p2){    return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));}point intersection(point u1,point u2,point v1,point v2){    point ret=u1;    double t=((u1.x-v1.x)*(v1.y-v2.y)-(u1.y-v1.y)*(v1.x-v2.x))/((u1.x-u2.x)*(v1.y-v2.y)-(u1.y-u2.y)*(v1.x-v2.x));    ret.x+=(u2.x-u1.x)*t;    ret.y+=(u2.y-u1.y)*t;    return ret;}int intersect_circle_circle(point c1,double r1,point c2,double r2){    return dis(c1,c2)<r1+r2+eps&&dis(c1,c2)>fabs(r1-r2)-eps;}void intersection_line_circle(point c,double r,point l1,point l2,point &p1,point &p2){    point p=c;    double t;    p.x+=l1.y-l2.y;    p.y+=l2.x-l1.x;    p=intersection(p,c,l1,l2);    t=sqrt(r*r-dis(p,c)*dis(p,c))/dis(l1,l2);    p1.x=p.x+(l2.x-l1.x)*t;    p1.y=p.y+(l2.y-l1.y)*t;    p2.x=p.x-(l2.x-l1.x)*t;    p2.y=p.x-(l2.y-l1.y)*t;}void intersection_circle_circle(point c1,double r1,point c2,double r2,point &p1,point &p2){    point u,v;    double t;    t=(1+(r1*r1-r2*r2)/dis(c1,c2)/dis(c1,c2))/2;    u.x=c1.x+(c2.x-c1.x)*t;    u.y=c1.y+(c2.y-c1.y)*t;    v.x=u.x+c1.y-c2.y;    v.y=u.y-c1.x+c2.x;    intersection_line_circle(c1,r1,u,v,p1,p2);}int main(){    int n;    while(scanf("%d",&n)!=EOF)    {        int cnt=0;        v.clear();        for(int i=0;i<n;i++)        {            cin>>c[i].p.x>>c[i].p.y>>c[i].r;            v.push_back(c[i].p);        }        for(int i=0;i<n;i++)            for(int j=0;j<i;j++)                if(intersect_circle_circle(c[i].p,c[i].r,c[j].p,c[j].r))                {                    point x,y;                    intersection_circle_circle(c[i].p, c[i].r, c[j].p, c[j].r, x, y);                    v.push_back(x);v.push_back(y);                }        int maxt=-1;        for(int i=0;i<v.size();i++)        {            int cnt=0;            for(int j=0;j<n;j++)                cnt+=dis(v[i],c[j].p)<c[j].r+eps;            maxt=max(cnt,maxt);        }        cout<<maxt<<endl;    }    return 0;}