BZOJ 2732: [HNOI2012]射箭
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Description
问一条过原点的抛物线最多能连续穿过几条线段.\(n \leqslant 10^5\)
Solution
二分+计算几何半平面交
过一条线段可以变成两个不等式,都写成\(ax+by+c\geqslant 0\)的形式.
这题蜜汁精度..
Code
/************************************************************** Problem: 2732 User: BeiYu Language: C++ Result: Accepted Time:5540 ms Memory:36028 kb ****************************************************************/ #include <bits/stdc++.h> using namespace std; namespace CG { typedef long double LD; const LD Pi = M_PI; const LD PI = 2 * acos(0.0); const LD eps = 1e-18; const LD oo = 1e15; #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; LD ang; Line(Point _p=Point(),Vector _v=Vector()):p(_p),v(_v) { ang=get_a(v); } 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; } int chkleft(Point P) { return dcmp(Cross(v,P-p))>0; } }; int cmpa(const Line &a,const Line &b) { return dcmp(a.ang-b.ang)==-1; } Point get_l_l(Line a,Line b) { Vector u=a.p-b.p; LD t=Cross(b.v,u)/Cross(a.v,b.v); return a.get_p(t); } typedef Line Hp; int get_h_h(vector<Hp> &hs,vector<Point> &pt) { sort(hs.begin(),hs.end(),cmpa); /* for(int i=0;i<(int)hs.size();i++) { hs[i].p.out();cout<<" ";hs[i].v.out();cout<<" "<<hs[i].ang<<endl; }*/ vector<Point> p(hs.size()); vector<Hp> q(hs.size()); int h,t; q[h=t=0]=hs[0]; /* for(int i=0;i<(int)hs.size();i++) { for(int j=0;j<(int)hs.size();j++) get_l_l(hs[i],hs[j]).out(),cout<<" "; cout<<endl; }*/ for(int i=1;i<(int)hs.size();i++) { while(h<t && !hs[i].chkleft(p[t-1])) t--; while(h<t && !hs[i].chkleft(p[h])) h++; q[++t]=hs[i]; if(fabs(Cross(q[t].v,q[t-1].v))<eps) { t--; if(q[t].chkleft(hs[i].p)) q[t]=hs[i]; } if(h<t) p[t-1]=get_l_l(q[t-1],q[t]); } while(h<t && !q[h].chkleft(p[t-1])) t--; // cout<<t-h+1<<endl; p[t]=get_l_l(q[h],q[t]); // for(int i=h;i<=t;i++) pt.push_back(p[i]); return t-h+1; } 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; const int N = 2e5+500; int n; LD xx[N],sy[N],ty[N]; vector<Line> ls; vector<Point> pt; int chk(int x) { ls.clear(); /* ls.push_back(Line(Point(0,0),Point(oo,0))); ls.push_back(Line(Point(-1e-13,0),Point(0,oo))); ls.push_back(Line(Point(0,oo),Point(-oo,0))); ls.push_back(Line(Point(-oo,0),Point(0,-oo)));*/ ls.push_back(Line(Point(0,0),Point(0,oo))); ls.push_back(Line(Point(0,oo),Point(-oo,0))); ls.push_back(Line(Point(-oo,oo),Point(0,-oo))); ls.push_back(Line(Point(-oo,0),Point(oo,0))); for(int i=1;i<=x;i++) { LD a=sqr(xx[i]),b=xx[i],c1=sy[i],c2=ty[i]; ls.push_back(Line(Point(0,c2/b),Point(-1e3,1e3*a/b))); ls.push_back(Line(Point(0,c1/b),Point(1e3,-1e3*a/b))); }return get_h_h(ls,pt)>=3; } int main() { scanf("%d",&n); for(int i=1;i<=n;i++) { int a,b,c; scanf("%d%d%d",&a,&b,&c); xx[i]=a,sy[i]=b-eps,ty[i]=c+eps; } // for(int i=1;i<=n;i++) cout<<chk(i)<<endl; int l=0,r=n,md; for(;l<=r;) { md=(l+r)>>1; if(chk(md)) l=md+1; else r=md-1; }printf("%d\n",r); return 0; }
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