註意等邊三角形的上頂點是卡不到邊界上的。

於是整個凸包分成三部分：左邊的連續的三角形、中間的、右邊的連續的三角形。

套個計算幾何板子求個三角形頂點到圓的切線、三角形頂點到正方形左上角距離啥的就行了，分類比較多。

#include<cstdio>
#include<cmath>
using namespace std;
const double PI=acos(-1.0);
int n;
char a[25];
struct Point{
	double x,y;
	double length(){
		return sqrt(x*x+y*y);
	}
};
typedef Point Vector;
Vector operator - (const Point &a,const Point &b){
	return (Vector){a.x-b.x,a.y-b.y};
}
Vector operator + (const Vector &a,const Vector &b){
	return (Vector){a.x+b.x,a.y+b.y};
}
Vector operator * (const double &K,const Vector &v){
	return (Vector){K*v.x,K*v.y};
}
Vector Rotate(Vector A,double rad)
{
	return (Vector){A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad)};
}
Vector unit(Vector A){
	double l=A.length();
	return (Vector){A.x/l,A.y/l};
}
double sqr(double x){
	return x*x;
}
int main(){
	scanf("%d%s",&n,a+1);
	bool alltr=1;
	for(int i=1;i<=n;++i){
		if(a[i]!=‘T‘){
			alltr=0;
			break;
		}
	}
	if(alltr){
		printf("%d\n",2*n+1);
	}
	else{
		double ans=0;
		int I,J;
		for(int i=1;i<=n;++i){
			if(a[i]!=‘T‘){
				if(i!=1){
					if(a[i]==‘S‘){
						ans+=((Point){0.5,sqrt(3.0)/2.0}-(Point){(double)(i-1),1.0}).length();
					}
					else if(a[i]==‘C‘){
						Vector v=(Point){(double)i-0.5,0.5}-(Point){0.5,sqrt(3.0)/2.0};
						ans+=sqrt(sqr(v.length())-0.5*0.5);
						v=Rotate(v,asin(0.5/v.length()));
						Point p=(Point){0.5,sqrt(3.0)/2.0}+ans*unit(v);
						double xian=((Vector){(double)i-0.5,1.0}-p).length();
						double jiao=acos((sqr(xian)-0.5*0.5*2.0)/(-2.0*0.5*0.5));
						ans+=jiao*0.5;
					}
				}
				else{
					if(a[i]==‘S‘){
						ans+=2.0;
					}
					else if(a[i]==‘C‘){
						ans+=PI*0.5;
					}
				}
				I=i;
				break;
			}
		}
		for(int j=n,i=1;j>=1;++i,--j){
			if(a[j]!=‘T‘){
				if(j!=n){
					if(a[j]==‘S‘){
						ans+=((Point){0.5,sqrt(3.0)/2.0}-(Point){(double)(i-1),1.0}).length();
					}
					else if(a[j]==‘C‘){
						Vector v=(Point){(double)i-0.5,0.5}-(Point){0.5,sqrt(3.0)/2.0};
						double d=sqrt(sqr(v.length())-0.5*0.5);
						ans+=d;
						v=Rotate(v,asin(0.5/v.length()));
						Point p=(Point){0.5,sqrt(3.0)/2.0}+d*unit(v);
						double xian=((Vector){(double)i-0.5,1.0}-p).length();
						double jiao=acos((sqr(xian)-0.5*0.5*2.0)/(-2.0*0.5*0.5));
						ans+=jiao*0.5;
					}
				}
				else{
					if(a[j]==‘S‘){
						ans+=2.0;
					}
					else if(a[j]==‘C‘){
						ans+=PI*0.5;
					}
				}
				J=j;
				break;
			}
		}
		if(I!=1 && J!=n){
			ans+=((double)(I-1)+1.0);
			ans+=((double)(n-J)+1.0);
			ans+=(double)(J-I+1);
			if(a[I]==‘S‘ && a[J]==‘S‘){
				ans+=(double)(J-I+1);
			}
			else if(a[I]==‘C‘ && a[J]==‘C‘){
				ans+=(double)(J-I);
			}
			else{
				ans+=((double)(J-I)+0.5);
			}
		}
		else if(I==1 && J==n){
			ans+=(double)(n-1)*2.0;
		}
		else if(I==1 && J!=n){
			ans+=((double)(n-J)+1.0);
			if(a[J]==‘S‘){
				ans+=((double)J-0.5)*2.0;
			}
			else{
				ans+=(((double)J-0.5)*2.0-0.5);
			}
			
		}
		else{
			ans+=((double)(I-1)+1.0);
			if(a[I]==‘S‘){
				ans+=((double)(n-I+1)-0.5)*2.0;
			}
			else{
				ans+=(((double)(n-I+1)-0.5)*2.0-0.5);
			}
		}
		printf("%.11f\n",ans);
	}
	return 0;
}

【計算幾何】【分類討論】Gym - 101173C - Convex Contour