[SHOI2012]信用卡凸包（凸包+直覺）

這個題仍是比較有趣。
當心發現，大膽猜測，不用證實！
咱們發現所謂的信用卡凸包上弧的長度總和就是圓的周長！
而後再加上每一個長寬都減去圓的直徑以後的長方形的凸包周長便可！

#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;
}