# BZOJ 1043: [HAOI2008]下落的圆盘

## Code

/**************************************************************
Problem: 1043
User: BeiYu
Language: C++
Result: Accepted
Time:520 ms
Memory:1308 kb
****************************************************************/

#include <bits/stdc++.h>
using namespace std;

namespace CG {
typedef double LD;

const LD Pi = M_PI;
const LD PI = 2 * acos(0.0);
const LD eps = 1e-12;
#define sqr(x) ((x)*(x))

int dcmp(LD x) { return fabs(x)<eps?0:(x<0?-1:1); }

struct Point {
LD x,y;
Point(LD _x=0,LD _y=0) :x(_x),y(_y) {}
void out() { cout<<"("<<x<<","<<y<<")"; }
};
typedef Point Vector;

int cmpx(const Point &a,const Point &b) { return dcmp(a.x-b.x)==0?a.y<b.y:a.x<b.x; }

Vector operator + (const Vector &a,const Vector &b) { return Vector(a.x+b.x,a.y+b.y); }
Vector operator - (const Vector &a,const Vector &b) { return Vector(a.x-b.x,a.y-b.y); }
Vector operator * (const Vector &a,LD b) { return Vector(a.x*b,a.y*b); }
Vector operator / (const Vector &a,LD b) { return Vector(a.x/b,a.y/b); }
bool operator == (const Point &a,const Point &b) { return dcmp(a.x-b.x)==0&&dcmp(a.y-b.y)==0; }

LD Dot(Vector a,Vector b) { return a.x*b.x+a.y*b.y; }
LD Cross(Vector a,Vector b) { return a.x*b.y-b.x*a.y; }
Vector Rot(Vector a,LD rd) { return Vector(a.x*cos(rd)-a.y*sin(rd),a.x*sin(rd)+a.y*cos(rd)); }
LD get_l(Vector a) { return sqrt(Dot(a,a)); }
LD get_d(Point a,Point b) { return sqrt(Dot(a-b,a-b)); }
LD get_a(Vector a) { return atan2(a.y,a.x); }
LD get_a(Vector a,Vector b) { return acos(Dot(a,b)/get_l(a)/get_l(b)); }
LD get_s(Point a,Point b,Point c) { return Cross(b-a,c-a)/2.0; }

struct Line {
Point p;
Vector v;
Line(Point a=Point(),Point b=Point()):p(a),v(b-a) {  }
LD get_l() { return sqrt(Dot(v,v)); }
Point get_p(LD t) { return p+v*t; }
Point get_s() { return p; }
Point get_t() { return p+v; }
};

struct Circle {
Point c;
LD r;
Point get_p(LD t) { return c+Point(cos(t)*r,sin(t)*r); }
LD get_rd(Point a,Point b) { return get_a(a-c,b-c); }
LD get_l(LD rd) { return r*rd; }
};

int get_c_l(Line L,Circle C,vector<Point> &res) {
LD a=L.v.x,b=L.p.x-C.c.x,c=L.v.y,d=L.p.y-C.c.y;
LD e=sqr(a)+sqr(c),f=2.0*(a*b+c*d),g=sqr(b)+sqr(d)-sqr(C.r);
LD dt=f*f-4*e*g;
if(dcmp(dt)<0) return 0;
if(dcmp(dt)==0) return res.push_back(L.get_p(-f/(2.0*e))),1;
LD x1=(-f-sqrt(dt))/(2.0*e),x2=(-f+sqrt(dt))/(2.0*e);
if(x1>x2) swap(x1,x2);
res.push_back(L.get_p(x1)),res.push_back(L.get_p(x2));return 2;
}
int get_c_c(Circle A,Circle B,vector<Point> &res) {
LD d=get_l(A.c-B.c);
if(dcmp(d)==0) return dcmp(A.r-B.r)==0?-1:0;
if(dcmp(A.r+B.r-d)<0) return 0;
if(dcmp(fabs(A.r-B.r)-d)>0) return 0;

LD a=get_a(B.c-A.c);
LD rd=acos((sqr(A.r)+sqr(d)-sqr(B.r))/(2.0*A.r*d));

Point p1,p2;
p1=A.get_p(a+rd),p2=A.get_p(a-rd);

res.push_back(p1);
if(p1==p2) return 1;
res.push_back(p2);
return 2;
}

/*---io---*/
ostream & operator << (ostream &os,const Point &p) { os<<p.x<<" "<<p.y;return os; }
istream & operator >> (istream &is,Point &p) { is>>p.x>>p.y;return is; }
ostream & operator << (ostream &os,const Circle &C) { os<<C.c<<" "<<C.r;return os; }
istream & operator >> (istream &is,Circle &C) { is>>C.c>>C.r;return is; }
};

using namespace CG;

#define mpr make_pair

int n;
LD ans;
vector<Circle> cr;
vector<Line> cl;
vector<Point> ls;
vector<Point> cp;

void add_l(vector<Point> &s,LD x,LD y) {
if(x>y) s.push_back(Point(x,Pi)),s.push_back(Point(-Pi,y));
else ls.push_back(Point(x,y));
}
int chk(Circle A,Circle B) {
LD d=get_l(A.c-B.c);
if(dcmp(B.r-A.r-d)>0) return 1;
return 0;
}
LD get_ans(Circle c,int id) {
ls.clear();
//O-O jiaodian
for(int i=id+1;i<n;i++) {
cp.clear();
if(chk(c,cr[i])) return 0;
if(get_c_c(c,cr[i],cp)<2) continue;
LD xx=get_a(cp[0]-c.c),yy=get_a(cp[1]-c.c);
}
if(!ls.size()) return c.get_l(2*Pi);
sort(ls.begin(),ls.end(),cmpx);
//  for(int i=0;i<(int)ls.size();i++) cout<<ls[i].x<<" "<<ls[i].y<<endl;
LD lx=ls[0].x,ly=lx,res=0;
for(int i=0;i<(int)ls.size();i++) {
if(dcmp(ls[i].x-ly)>0) res+=ly-lx,lx=ls[i].x;
ly=max(ly,ls[i].y);
}res+=ly-lx;
return c.get_l(2*Pi-res);
}
void Solve() {
scanf("%d",&n);
Circle cc;
cr.clear(),ls.clear(),cp.clear(),ans=0;

for(int i=1;i<=n;i++) {
scanf("%lf%lf%lf",&cc.r,&cc.c.x,&cc.c.y);
cr.push_back(cc);
}
for(int i=0;i<n;i++) {
LD tmp=get_ans(cr[i],i);
ans+=tmp;
//  cout<<tmp<<endl;
}printf("%.3lf\n",ans);
}

int main() {
Solve();
return 0;
}


posted @ 2017-04-17 08:53  北北北北屿  阅读(116)  评论(0编辑  收藏  举报