[SHOI2012]信用卡凸包（凸包+直觉）

这个题仍是比较有趣。
当心发现，大胆猜测，不用证实！
咱们发现所谓的信用卡凸包上弧的长度总和就是圆的周长！
而后再加上每一个长宽都减去圆的直径以后的长方形的凸包周长便可！

#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;
}