bzoj2829 信用卡凸包

Posted 2020-07-08 AaronPolaris

2829: 信用卡凸包

Time Limit: 10 Sec  Memory Limit: 128 MBSec  Special Judge
Submit: 226  Solved: 102
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Description

Input

Output

Sample Input

2
6.0 2.0 0.0
0.0 0.0 0.0
2.0 -2.0 1.5707963268

Sample Output

21.66

HINT

本样例中的2张信用卡的轮廓在上图中用实线标出，如果视1.5707963268为

Pi/2(pi为圆周率),则其凸包的周长为16+4*sqrt(2)

凸包

将图形去掉外面一层，留下中间的矩形，求出凸包，最后加上一个圆的周长。

#include<iostream>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<cstdio>
#include<algorithm>
#define F(i,j,n) for(int i=j;i<=n;i++)
#define D(i,j,n) for(int i=j;i>=n;i--)
#define ll long long
#define maxn 400005
#define eps 1e-8
using namespace std;
int n,cnt,top;
double a,b,r,x,y,angle,ans;
const double pi=acos(-1);
struct P
{
	double x,y;
	P(double xx=0,double yy=0){x=xx;y=yy;}
	friend P operator -(P a,P b){return P(a.x-b.x,a.y-b.y);}
	friend double operator *(P a,P b){return a.x*b.y-a.y*b.x;}
	friend bool operator <(P a,P b){return fabs(a.y-b.y)<eps?a.x<b.x:a.y<b.y;}
}p[maxn],s[maxn];
inline P move(P a,double len,double angle)
{
	return P(a.x+len*cos(angle),a.y+len*sin(angle));
}
inline double dis(P a,P b)
{
	return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);
}
inline bool cmp(P a,P b)
{
	double tmp=(a-p[1])*(b-p[1]);
	return fabs(tmp)<=eps?dis(p[1],a)<dis(p[1],b):tmp>0;
}
inline void solve()
{
	F(i,2,cnt) if (p[i]<p[1]) swap(p[1],p[i]);
	sort(p+2,p+cnt+1,cmp);
	F(i,1,cnt)
	{
		while (top>1&&(s[top]-s[top-1])*(p[i]-s[top-1])<eps) top--;
		s[++top]=p[i];
	}
}
int main()
{
	scanf("%d",&n);
	scanf("%lf%lf%lf",&a,&b,&r);
	a-=2*r;b-=2*r;ans=2*pi*r;
	F(i,1,n)
	{
		scanf("%lf%lf%lf",&x,&y,&angle);
		F(j,1,4)
		{
			P tmp=move(P(x,y),b/2,angle+pi/2*(j-1));
			p[++cnt]=move(tmp,a/2,angle+pi/2*j);
			swap(a,b);
		}
	}
	solve();
	s[top+1]=s[1];
	F(i,1,top) ans+=sqrt(dis(s[i],s[i+1]));
	printf("%.2lf\n",ans);
	return 0;
}