BZOJ 2829 凸包

思路：

把信用卡周围去掉  只剩下中间的长方形 

最后的答案加上一个圆

//By SiriusRen
#include <bits/stdc++.h>
using namespace std;
const int N=50000;
const double eps=1e-10;
int card,n,m,k;
double a,b,r,xx,yy,theta,Ans;
struct Point{
    double x,y;Point(){}
    Point(double X,double Y){x=X,y=Y;}
}point[N],tubao[N];
bool cmp1(Point a,Point b){if(abs(a.x-b.x)>eps)return a.x<b.x;return a.y<b.y;}
double cross(Point a,Point b,Point c){
    return (a.x-c.x)*(b.y-c.y)-(a.y-c.y)*(b.x-c.x);
}
double dis(Point a,Point b){
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
int main(){
    scanf("%d%lf%lf%lf",&card,&a,&b,&r);
    Ans+=2*acos(-1)*r;
    a=a-2*r,b=b-2*r;
    for(int i=1;i<=card;i++){
        scanf("%lf%lf%lf",&xx,&yy,&theta);
        point[++n]=Point(xx-sin(theta)*a/2-cos(theta)*b/2,yy+cos(theta)*a/2-sin(theta)*b/2);
        point[++n]=Point(xx-sin(theta)*a/2+cos(theta)*b/2,yy+cos(theta)*a/2+sin(theta)*b/2);
        point[++n]=Point(xx+sin(theta)*a/2-cos(theta)*b/2,yy-cos(theta)*a/2-sin(theta)*b/2);
        point[++n]=Point(xx+sin(theta)*a/2+cos(theta)*b/2,yy-cos(theta)*a/2+sin(theta)*b/2);
    }
    sort(point+1,point+1+n,cmp1);
    for(int i=1;i<=n;i++){
        while(m>1&&cross(tubao[m],point[i],tubao[m-1])<eps)m--;
        tubao[++m]=point[i];
    }k=m;
    for(int i=n-1;i;i--){
        while(m>k&&cross(tubao[m],point[i],tubao[m-1])<eps)m--;
        tubao[++m]=point[i];
    }
    for(int i=1;i<m;i++)Ans+=dis(tubao[i],tubao[i+1]);
    printf("%.2lf\n",Ans);
}