hdu 1077 （圆交）

Problem - 1077

　　我们可以知道，当这个单位圆可以覆盖到最多的点的时候，必定最少有两个点位于这个圆的圆周上，于是就有网上众多的O(N^3)的枚举两个在圆上的点的暴搜做法。

　　然而这题是可以用圆交来做的。

　　我们以一条鱼的位置作为圆心，半径为1的圆的周围随便找一个点都能把这条鱼抓到。这时，我们可以做出很多个这样的圆，半径都为1。

　　然后，求一下这些圆的交集，叠起来的最高层数就是最多能获得的鱼的数目。

　　这里的圆交不需要实现求面积这部分，于是只需要离散一下交点就行了。

　　1y。

 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 #include <iomanip>
 #include <cmath>

 using namespace std;

 const double EPS = 1e-;
 const double PI = acos(-1.0);
 inline int sgn(double x) { return (x > EPS) - (x < -EPS);}
 template<class T> T sqr(T x) { return x * x;}
 typedef pair<double, double> Point;
 #define x first
 #define y second

 Point operator + (Point a, Point b) { return Point(a.x + b.x, a.y + b.y);}
 Point operator - (Point a, Point b) { return Point(a.x - b.x, a.y - b.y);}
 Point operator * (Point a, double p) { return Point(a.x * p, a.y * p);}
 Point operator * (double p, Point a) { return Point(a.x * p, a.y * p);}
 Point operator / (Point a, double p) { return Point(a.x / p, a.y / p);}

 inline double cross(Point a, Point b) { return a.x * b.y - a.y * b.x;}
 inline double dot(Point a, Point b) { return a.x * b.x + a.y * b.y;}
 inline double veclen(Point a) { return sqrt(dot(a, a));}
 inline double angle(Point a) { return atan2(a.y, a.x);}
 inline Point vecunit(Point a) { return a / veclen(a);}
 inline Point normal(Point a) { return Point(-a.y, a.x) / veclen(a);}

 struct Line {
     Point s, t;
     Line() {}
     Line(Point s, Point t) : s(s), t(t) {}
     Point vec() { return t - s;}
     Point point(double p) { return s + vec() * p;}
 } ;
 inline Point llint(Line a, Line b) { return a.point(cross(b.vec(), a.s - b.s) / cross(a.vec(), b.vec()));}

 struct Circle {
     Point c;
     double r;
     Circle() {}
     Circle(Point c, double r) : c(c), r(r) {}
     Point point(double p) { return c + r * Point(cos(p), sin(p));}
     bool in(Point p) { return sgn(veclen(p - c) - r) < ;}
 } ;

 const double R = 1000.0;
 const int N = ;
 int n;
 Circle cir[N];

 void input() {
     cin >> n;
     for (int i = ; i < n; i++) {
         cin >> cir[i].c.x >> cir[i].c.y;
         cir[i].c = cir[i].c * R;
         cir[i].r = R;
     }
 }

 bool ccint(int _a, int _b, double *sol) {
     Circle a = cir[_a], b = cir[_b];
     double d = veclen(a.c - b.c), dr = fabs(a.r - b.r);
     if (sgn(d - a.r - b.r) > ) return ;
     if (sgn(dr - d) >= ) {
         if (a.r < b.r || sgn(a.r - b.r) ==  && _a < _b) {
             sol[] = -PI;
             sol[] = PI;
             return ;
         }
         return ;
     }
     double ang = angle(b.c - a.c);
     double da = acos(fabs(sqr(a.r) + sqr(d) - sqr(b.r)) / ( * a.r * d));
     sol[] = ang - da;
     sol[] = ang + da;
     return ;
 }

 typedef pair<double, int> Event;
 Event ev[N << ];
 bool cmp(Event a, Event b) {
     if (sgn(a.x - b.x)) return a.x < b.x;
     return a.y > b.y;
 }

 int cal(int id) {
     int tt = ;
     double _[];
     for (int i = ; i < n; i++) {
         if (i == id) continue;
         if (ccint(id, i, _)) {
             if (_[] < -PI) {
                 ev[tt++] = Event(_[] +  * PI, );
                 ev[tt++] = Event(PI, -);
                 ev[tt++] = Event(-PI, );
                 ev[tt++] = Event(_[], -);
             } else if (_[] > PI) {
                 ev[tt++] = Event(_[], );
                 ev[tt++] = Event(PI, -);
                 ev[tt++] = Event(-PI, );
                 ev[tt++] = Event(_[] -  * PI, -);
             } else {
                 ev[tt++] = Event(_[], );
                 ev[tt++] = Event(_[], -);
             }
         }
     }
     int cnt = , mx = ;
     sort(ev, ev + tt, cmp);
     for (int i = ; i < tt; i++) {
         cnt += ev[i].y;
         mx = max(mx, cnt);
     }
     return mx;
 }

 int work() {
     int mx = ;
     for (int i = ; i < n; i++) mx = max(mx, cal(i));
     return mx;
 }

 int main() {
     //freopen("in", "r", stdin);
     ios::sync_with_stdio();
     cout << setiosflags(ios::fixed) << setprecision();
     int _;
     cin >> _;
     while (_--) {
         input();
         cout << work() << endl;
     }
     return ;
 }

——written by Lyon