hdu 3264 Open-air shopping malls 求兩圓相交

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對每個圓二分半徑尋找可行的最小半徑，然後取最小的一個半徑。

對於兩圓相交就只要求到兩個扇形，然後減去兩個全等三角形就行了。

#include<cstdio>#include<iostream>#include<cmath>#include<algorithm>using namespace std;#define pi acos(-1.0)#define eps 1e-8#define maxn 50int n;struct point{    double x;    double y;    double r;}c[maxn];double dis(point a,point b){   return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}double area(point a,double ra,point b,double rb){    double ans=0;    double d=dis(a,b);    double temp;    if(ra<rb)  swap(ra,rb);    if(d>=ra+rb)return 0;    if(d<=ra-rb)return pi*rb*rb;    double angle1=acos((ra*ra+d*d-rb*rb)/2.0/ra/d);    double angle2=acos((rb*rb+d*d-ra*ra)/2.0/rb/d);    ans-=d*ra*sin(angle1);    ans+=angle1*ra*ra+angle2*rb*rb;    return ans;}bool cover_half(point a,double ra,point b,double rb)  //a是有傘的圓，b是其他圓{    return area(a,ra,b,rb)>=0.5*rb*rb*pi;}bool isok(double r,int k){    for(int i=1;i<=n;i++)    {        if(!cover_half(c[k],r,c[i],c[i].r)) return false;    }    return true;}int main(){    int cas;    scanf("%d",&cas);    while(cas--)    {        scanf("%d",&n);        for(int i=1;i<=n;i++)        {            scanf("%lf%lf%lf",&c[i].x,&c[i].y,&c[i].r);        }        double ans=5000000;        for(int i=1;i<=n;i++){            double l=0.0,r=5000000,mid;            while(l+eps<=r)            {                mid=(l+r)/2;                if(isok(mid,i)) r=mid-eps;                else l=mid+eps;            }            ans=min(ans,mid);        }        printf("%.4lf\n",ans);    }    return 0;}