BZOJ 1043: [HAOI2008]下落的圆盘
Description
求几个圆交起来的周长..n⩽103
Solution
计算几何.
圆圆求交..
Code
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 | /************************************************************** 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); add_l(ls,yy,xx); } 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; } |
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