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  • 【计算几何】【凸包】bzoj2829 信用卡凸包

    http://hzwer.com/6330.html

    #include<cstdio>
    #include<cmath>
    #include<algorithm>
    using namespace std;
    #define N 100001
    #define EPS 0.00000001
    typedef double db;
    const db PI=acos(-1.0);
    struct Point{db x,y;}p[N<<2],bao[N<<2];
    bool operator < (Point a,Point b){return fabs(a.x-b.x)>=EPS?a.x<b.x:a.y<b.y;}
    typedef Point Vector;
    Vector operator + (Vector a,Vector b){return (Vector){a.x+b.x,a.y+b.y};}
    Vector operator - (Vector a,Vector b){return (Vector){a.x-b.x,a.y-b.y};}
    Vector Rotate(Vector a,db rad)
    {return (Vector){a.x*cos(rad)-a.y*sin(rad),a.x*sin(rad)+a.y*cos(rad)};}
    db sqr(db x){return x*x;}
    db Cross(Vector a,Vector b){return a.x*b.y-a.y*b.x;}
    db dist(Point a,Point b){return sqrt(sqr(a.x-b.x)+sqr(a.y-b.y));}
    int n,m,en;
    db a,b,r,mx,my,jiao,ans;
    int main()
    {
    	scanf("%d%lf%lf%lf",&n,&b,&a,&r);
    	a-=(2.0*r);
    	b-=(2.0*r);
    	for(int i=1;i<=n;++i)
    	  {
    	  	scanf("%lf%lf%lf",&mx,&my,&jiao);
    	  	p[++m]=Rotate((Vector){a*0.5,b*0.5},jiao)+(Point){mx,my};
    	  	p[++m]=Rotate((Vector){(-a)*0.5,b*0.5},jiao)+(Point){mx,my};
    	  	p[++m]=Rotate((Vector){a*0.5,(-b)*0.5},jiao)+(Point){mx,my};
    	  	p[++m]=Rotate((Vector){(-a)*0.5,(-b)*0.5},jiao)+(Point){mx,my};
    	  }
    	sort(p+1,p+1+m);
    	for(int i=1;i<=m;++i)
    	  {
    	  	while(en>1&&Cross(bao[en]-bao[en-1],p[i]-bao[en])<=0.0)
    	  	  --en;
    	  	bao[++en]=p[i];
    	  }
    	int t=en;
    	for(int i=m-1;i;--i)
    	  {
    	  	while(en>t&&Cross(bao[en]-bao[en-1],p[i]-bao[en])<=0.0)
    	  	  --en;
    	  	bao[++en]=p[i];
    	  }
    	for(int i=2;i<=en;++i)
    	  ans+=dist(bao[i-1],bao[i]);
    	printf("%.2f
    ",ans+dist(bao[en],bao[1])+PI*r*2.0);
    	return 0;
    }
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  • 原文地址:https://www.cnblogs.com/autsky-jadek/p/4378657.html
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