Codeforces gym102222 B.Rolling The Polygon 凸包/余弦定理

题意：

有一个不保证凸的多边形，让你滚一圈，计算某点滚出的轨迹多长。

题解：

求出凸包后，以每个点为转轴，转轴到定点的距离为半径，用余弦定理计算圆心角，计算弧长。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
using namespace std;
int main()
{
    int t,case1=;
    cin>>t;
    while(t--)
    {
        case1++;
        int n;
        cin>>n;
        int x[],y[];//贮存坐标
        for(int i=;i<=n;i++)
        {
            cin>>x[i]>>y[i];
        }
        int xx,yy;
        cin>>xx>>yy;
        double dis[];
        for(int i=;i<=n;i++)
        {
            dis[i]=(x[i]-xx)*(x[i]-xx)+(y[i]-yy)*(y[i]-yy);
            //cout<<dis[i]<<endl;
        }
        //cout<<endl;
        double ankle[];
        for(int i=;i<=n;i++)
        {
            double t1,t2,t3;
            if(xx==x[i]&&yy==y[i])
            continue;
            if(i==)
            {
                t1=sqrt((x[]-x[])*(x[]-x[])+(y[]-y[])*(y[]-y[]));
                t2=sqrt((x[n]-x[])*(x[n]-x[])+(y[n]-y[])*(y[n]-y[]));
                t3=sqrt((x[n]-x[])*(x[n]-x[])+(y[n]-y[])*(y[n]-y[]));
            }
            else if(i==n)
            {
                t1=sqrt((x[n]-x[])*(x[n]-x[])+(y[n]-y[])*(y[n]-y[]));
                t2=sqrt((x[n]-x[n-])*(x[n]-x[n-])+(y[n]-y[n-])*(y[n]-y[n-]));
                t3=sqrt((x[n-]-x[])*(x[n-]-x[])+(y[n-]-y[])*(y[n-]-y[]));
            }
            else
            {
                t1=sqrt((x[i+]-x[i])*(x[i+]-x[i])+(y[i+]-y[i])*(y[i+]-y[i]));
                t2=sqrt((x[i]-x[i-])*(x[i]-x[i-])+(y[i]-y[i-])*(y[i]-y[i-]));
                t3=sqrt((x[i-]-x[i+])*(x[i-]-x[i+])+(y[i-]-y[i+])*(y[i-]-y[i+]));
            }
            //cout<<t1<<" "<<t2<<" "<<t3<<" ";
            ankle[i]=acos((t1*t1+t2*t2-t3*t3)/(*t1*t2));
            //cout<<ankle[i]<<" ";
            //cout<<endl;
        }

        double l=;
        for(int i=;i<=n;i++)
        {
            l+=sqrt(dis[i])*(acos(-1.0)-ankle[i]);
        }
        printf("Case #%d: %.3lf\n",case1,l);
    }
    return ;
}