计算几何专项：UVa 10969

这道题和训练指南上的例题LA2572几乎一样，唯一不同的是只是求可见的周长。具体就是把所有的圆分成一段一段的圆弧（用圆弧的中点代表圆弧），然后看圆上的每一段圆弧是否在顶部，具体的判别方法需要一些技巧，因为精度问题在这上面WA了一次。具体参见代码吧。

#include <iostream>#include <cstdio>#include <cmath>#include <vector>#include <algorithm>using namespace std;#define M 110const double eps=1e-10;const double pi=acos(-1.0);int dcmp(double x){    if(fabs(x)<eps) return 0;    else return x<0?-1:1;}struct point{    double x,y;    point(double x=0,double y=0):x(x),y(y){}};point operator-(point a,point b){return point(a.x-b.x,a.y-b.y);}double dot(point a,point b){return a.x*b.x+a.y*b.y;}double length(point a){return sqrt(dot(a,a));}double cross(point a,point b){return a.x*b.y-a.y*b.x;}double angle(point v){return atan2(v.y,v.x);}double normal(double rad){    return rad-2*pi*floor(rad/(2*pi));}void getcircleinter(point c1,double r1,point c2,double r2,vector<double>& rad){    double d=length(c1-c2);    if(dcmp(d)==0) return;    if(dcmp(r1+r2-d)<0) return;    if(dcmp(fabs(r1-r2)-d)>0) return;    double a=angle(c2-c1);    double da=acos((r1*r1+d*d-r2*r2)/(2*r1*d));    rad.push_back(normal(a+da));    rad.push_back(normal(a-da));}int n;point center[M];double radius[M];int topmost(point p){    for(int i=n-1;i>=0;i--)        if(dcmp(length(center[i]-p)-radius[i])<0) return i;    return -1;}int main(){    freopen("in.txt","r",stdin);    int T;    cin>>T;    while(T--)    {        cin>>n;        for(int i=0;i<n;i++) cin>>radius[i]>>center[i].x>>center[i].y;        double ans=0;        for(int i=0;i<n;i++)        {            vector<double> rad;            rad.push_back(0);            rad.push_back(2*pi);            for(int j=0;j<n;j++)            {                getcircleinter(center[i],radius[i],center[j],radius[j],rad);            }            sort(rad.begin(),rad.end());            for(int j=0;j<rad.size()-1;j++)            {                double mid=(rad[j]+rad[j+1])/2.0;                int flag=0;                for(int side=-2;side<=2;side+=4)                {                    double r=radius[i]+side*eps;                    int tmp=topmost(point(center[i].x+r*cos(mid),center[i].y+r*sin(mid)));                    if(tmp==i) flag=1;                }                if(flag) ans+=(rad[j+1]-rad[j])*radius[i];            }        }        printf("%.3f\n",ans);    }    return 0;}