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BZOJ2829信用卡凸包

2015-06-02 13:11 645 查看
将每个信用卡内小矩形的四个点加进点集,求凸包,在加上pi*r*r。

证明原理是:凸多边形外角和等于2Pi rad.

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<queue>
#include<cmath>
#include<vector>
using namespace std;
#define Vector Point
const double pi=3.1415926535898,eps=1e-6;
int n,sz=0,cur=0;
double a,b,r,l,ans;
struct Point{ double x,y; Point(double x=0,double y=0):x(x),y(y){} }p[610000],q1[610000],q2[610000];
void Wa(){ cout<<"Wrong!!!"<<endl; system("pause"); }
void in(Point &u){ scanf("%lf%lf",&u.x,&u.y); }
double dcmp(double x){ if(abs(x)<eps)return 0; return x; }
int Getsign(double x){ if(!dcmp(x))return 0; if(x<0)return -1; return 1; }
bool operator <(Point u, Point v){ return u.x<v.x||(v.x==u.x&&u.y<v.y); }
bool operator ==(Point u, Point v){ return abs(u.x-v.x)<eps&&abs(u.y-v.y)<eps; }
bool operator >(Point u, Point v){ return !(u==v||u<v); }
Vector operator -(Point u, Point v){ return Point(u.x-v.x,u.y-v.y); }
Vector operator +(Vector u, Vector v){ return Vector(u.x+v.x,u.y+v.y); }
double operator *(Vector u, Vector v){ return u.x*v.x+u.y*v.y; }
Vector operator *(Vector u, double k){ return Vector(u.x*k, u.y*k); }
Vector operator /(Vector u, double k){ if(!k)Wa(); return Vector(u.x/k, u.y/k); }
double Cross(Vector u, Vector v){ return u.x*v.y-u.y*v.x; }
double AreaTri(Vector u, Vector v){ return Cross(u,v); }
Vector Rotate(Vector u, double rad){ return Vector(u.x*cos(rad)-u.y*sin(rad), u.x*sin(rad)+u.y*cos(rad)); }
double Length(Vector u){ return sqrt(u.x*u.x+u.y*u.y); }
double Angle(Vector u, Vector v){ if(!(Length(u)&&Length(v)))Wa(); return acos(u*v/Length(u)/Length(v)); }
Point Getint(Point a, Vector b, Point c, Vector d){
Vector tem=a-c;  double t=Cross(d,tem)/Cross(b,d); return a+b*t; }
bool Onseg(Point a,Point b,Point c){ return dcmp((a-c)*(b-c))==0&&dcmp(Cross(a-c,b-c))<0; }
bool IntLine(Point a, Point b, Point c, Point d)
{
double c1=Cross(b-a,c-a),c2=Cross(b-a,d-a);
double c3=Cross(c-d,a-d),c4=Cross(c-d,b-d);
return Getsign(c1)*Getsign(c2)<0 &&Getsign(c3)*Getsign(c4)<0;
}
bool cmpx(Point u, Point v){ return (v.x-u.x)>eps||(abs(u.x-v.x)<eps&&u.y<v.y); }
bool cmp (Point u, Point v){ return u==v; }
int main ()
{
scanf("%d%lf%lf%lf",&n,&a,&b,&r);
for(int i=1;i<=n;i++){
double x,y,z;
scanf("%lf%lf%lf",&x,&y,&z); Vector tem=Vector(x,y);
p[++sz]=tem+Rotate(Vector(b/2-r,a/2-r),z);
p[++sz]=tem+Rotate(Vector(-b/2+r,a/2-r),z);
p[++sz]=tem+Rotate(Vector(b/2-r,-a/2+r),z);
p[++sz]=tem+Rotate(Vector(-b/2+r,-a/2+r),z);
}
sort(p+1,p+1+sz,cmpx); q1[++cur]=p[1],q1[++cur]=p[2];
for(int i=3;i<=sz;i++){
while(cur>1&&Cross(q1[cur]-q1[cur-1],p[i]-q1[cur])<=0)cur--;
q1[++cur]=p[i];
}
for(int i=1;i<cur;i++)ans+=Length(q1[i+1]-q1[i]);
cur=0; q2[++cur]=p[1],q2[++cur]=p[2];
for(int i=3;i<=sz;i++){
while(cur>1&&Cross(q2[cur]-q2[cur-1],p[i]-q2[cur])>=0)cur--;
q2[++cur]=p[i];
}
for(int i=1;i<cur;i++)ans+=Length(q2[i+1]-q2[i]);
printf("%.2f\n",ans+2*r*pi);   return 0;
}

模版看起来有点恶心哈哈哈哈。

q1,q2中的点分别是下凸壳和上凸壳。
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标签:  计算几何 凸包