[SHOI2012]信用卡凸包（凸包+直觉）

这个题还是比较有趣。

小心发现，大胆猜想，不用证明！

我们发现所谓的信用卡凸包上弧的长度总和就是圆的周长！

然后再加上每个长宽都减去圆的直径之后的长方形的凸包周长即可！

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
const int N=40100;
const double eps=1e-12;
int stack[N],top,n;
double a,b,r,ans;
struct node{
	double x,y;
	node(double xx=0,double yy=0){
		x=xx;y=yy;
	}
}c[N];
node work(node a,double x){
	double A=cos(x),B=sin(x);
	return node(a.x*A-a.y*B,a.x*B+a.y*A);
}
bool cmp(node a,node b){
	if(a.x==b.x)return a.y<b.y;
	else return a.x<b.x;
}
double chaji(node a,node b){
	return a.x*b.y-a.y*b.x;
}
node operator -(node a,node b){
	return node(a.x-b.x,a.y-b.y);
}
node operator +(node a,node b){
	return node(a.x+b.x,a.y+b.y);
}
double dis(node a,node b){
	return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
int main(){
	scanf("%d",&n);
	scanf("%lf%lf%lf",&a,&b,&r);
	a-=2.0*r;b-=2.0*r;
	ans=acos(-1.0)*r*2.0;
	for(int i=0;i<n;i++){
		double x,y,z;
		scanf("%lf%lf%lf",&x,&y,&z);
		c[i*4+1].x=b/2.0;c[i*4+1].y=a/2.0;
		c[i*4+1]=work(c[i*4+1],z);c[i*4+1].x+=x;c[i*4+1].y+=y;
		c[i*4+2].x=-b/2.0;c[i*4+2].y=a/2.0;
		c[i*4+2]=work(c[i*4+2],z);c[i*4+2].x+=x;c[i*4+2].y+=y;
		c[i*4+3].x=b/2.0;c[i*4+3].y=-a/2.0;
		c[i*4+3]=work(c[i*4+3],z);c[i*4+3].x+=x;c[i*4+3].y+=y;
		c[i*4+4].x=-b/2.0;c[i*4+4].y=-a/2.0;
		c[i*4+4]=work(c[i*4+4],z);c[i*4+4].x+=x;c[i*4+4].y+=y;
	}
	sort(c+1,c+1+n*4,cmp);
	for(int i=1;i<=n*4;i++){
		if(top<=1){stack[++top]=i;continue;}
		while(top>=2&&chaji(c[stack[top]]-c[stack[top-1]],c[i]-c[stack[top]])+eps<0)top--;
		stack[++top]=i;
	}
	for(int i=1;i<top;i++)ans+=dis(c[stack[i]],c[stack[i+1]]);
	top=0;
	for(int i=n*4;i>=1;i--){
		if(top<=1){stack[++top]=i;continue;}
		while(top>=2&&chaji(c[stack[top]]-c[stack[top-1]],c[i]-c[stack[top]])+eps<0)top--;
		stack[++top]=i;
	}
	for(int i=1;i<top;i++)ans+=dis(c[stack[i]],c[stack[i+1]]);
	printf("%.2lf",ans);
	return 0;
}