Sicily8544

一条从原点出发的射线，求所有和它相交的圆的相交弦的和的最大值。

从Y轴开始共24*60个时刻，旋转一周。

每个时刻转的角度为360°/(24*60)。

直接一个一个枚举判断即可。

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <vector>#include <cmath>#include <algorithm>using namespace std;struct circle{double x,y,r;};struct circle c[205];int n;double cal(double k,double angle){int i;double ans=0;double pi=acos(-1);for (i=0;i<n;i++){double length2=(k*c[i].x-c[i].y)*(k*c[i].x-c[i].y)/(k*k+1);double xx=(c[i].x+k*c[i].y)/(k*k+1);double yy=k*xx;if (length2<=c[i].r*c[i].r){if (angle>pi/2 && angle<=pi && xx<0 && yy>=0)  ans+=sqrt(c[i].r*c[i].r-length2)*2;else if (angle>pi && angle<3*pi/2 && xx<0 && yy<0)ans+=sqrt(c[i].r*c[i].r-length2)*2;else if (angle>3*pi/2 && angle<=2*pi && xx>0 && yy<=0)ans+=sqrt(c[i].r*c[i].r-length2)*2;else if (angle>2*pi && xx>0 && yy>0)ans+=sqrt(c[i].r*c[i].r-length2)*2;}}return ans;}int main(){while (1){int i;scanf("%d",&n);if (!n) break;for (i=0;i<n;i++)scanf("%lf%lf%lf",&c[i].x,&c[i].y,&c[i].r);double per=2*acos(-1)/(24*60);double maxn=0;double ans=0;for (i=0;i<n;i++){if (c[i].r>fabs(c[i].x) && c[i].y>0)ans+=sqrt(c[i].r*c[i].r-c[i].x*c[i].x)*2;}maxn=ans;ans=0;for (i=0;i<n;i++){if (c[i].r>fabs(c[i].x) && c[i].y<0)ans+=sqrt(c[i].r*c[i].r-c[i].x*c[i].x)*2;}if (ans>maxn)maxn=ans;double angle=acos(-1)/2;for (i=1;i<=24*60-1;i++){angle+=per;double k=tan(angle);double temp=cal(k,angle);if (temp>maxn)maxn=temp;}printf("%.3lf\n",maxn);}}