Luogu-3829 [SHOI2012]信用卡凸包

这道题的转化很巧妙，可以把信用卡四个角的圆心看做平面上的点来做凸包，\(ans\)就是凸包周长加上一个圆的周长

// luogu-judger-enable-o2
#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=1e5+100;
const double Pi=3.14159265358979323846;
struct Point{
    double x,y;
    Point(double xx=0,double yy=0){
        x=xx,y=yy;
    }
    bool operator < (Point a) const{
        return x==a.x?y<a.y:x<a.x;
    }
    bool operator == (Point a) const{
        return x==a.x&&y==a.y;
    }
}a[maxn],b[maxn],c;
struct Vector{
    double x,y;
    Vector(double xx=0,double yy=0){
        x=xx,y=yy;
    }
}zhy[4];
int dcmp(double x){return fabs(x)<1e-9?0:(x>0?1:-1);}
Vector operator - (Point a,Point b){return Vector(a.x-b.x,a.y-b.y);}
Point operator + (Point a,Vector b){return Point(a.x+b.x,a.y+b.y);}
double operator * (Vector a,Vector b){return a.x*b.y-a.y*b.x;}
Vector rotate(Vector a,double p){return Vector(a.x*cos(p)-a.y*sin(p),a.x*sin(p)+a.y*cos(p));}
double dot(Vector a,Vector b){return a.x*b.x+a.y*b.y;}
double len(Vector a){return sqrt(dot(a,a));}
int n,m,tot;
double r,du;
void tb(Point *p,int n,Point *q,int &m){
    sort(p+1,p+n+1);
    q[m=1]=p[1];
    if(n==1) return;
    for(int i=2;i<=n;i++){
        while(m>1&&dcmp((q[m]-q[m-1])*(p[i]-q[m-1]))<=0)
            m--;
        q[++m]=p[i];
    }
    int k=m;
    for(int i=n-1;i>=1;i--){
        while(m>k&&dcmp((q[m]-q[m-1])*(p[i]-q[m-1]))<=0)
            m--;
        q[++m]=p[i];
    }
    m--;
}
double C(Point *p,int n){
    if(n==1) return 0;
    double tot=0;
    for(int i=1;i<n;i++)
        tot+=len(p[i+1]-p[i]);
    return tot+len(p[n]-p[1]);
}
int main(){
    scanf("%d",&m);
    scanf("%lf%lf%lf",&zhy[0].y,&zhy[0].x,&r);
    zhy[0].x/=2,zhy[0].y/=2;
    zhy[0].x-=r,zhy[0].y-=r;
    zhy[1]=zhy[2]=zhy[3]=zhy[0];
    zhy[1].x*=-1,zhy[2].x*=-1,zhy[2].y*=-1,zhy[3].y*=-1;
    for(int i=1;i<=m;i++){
        scanf("%lf%lf%lf",&c.x,&c.y,&du);
        for(int j=0;j<4;j++)
            a[++tot]=c+rotate(zhy[j],du);
    }
    tb(a,tot,b,n);
    double ans=C(b,n);
    ans=ans+2*Pi*r;
    printf("%.2lf\n",ans);
    return 0;
}