Sphere Online Judge (SPOJ) - Problem CIRU
【求圆并的若干种算法,圆并扩展算法】_AekdyCoin的空间_百度空间
参考AekdyCoin的圆并算法解释,根据理解写出的代码。圆并这么多题中,最基础一题。
操作如下:
(1)对一个圆,获得所有与其他圆的交点作为时间戳。
a.如果这个圆不被其他任何圆覆盖,这个圆直接保留。
b.如果Case中有两个完全重合的圆,可以保留标号小(或大)的圆,也只保留这一个。
(2)每经过一个时间戳就对计数器加减,如果计数器从0变1,加上这段圆弧和对应的三角形。
(3)所有这样的圆独立计算,最后求出的面积累加即可。
代码如下(1y):
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm> using namespace std; const double PI = acos(-1.0);
const double EPS = 1e-;
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 / (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 angle(Point a) { return atan2(a.y, a.x);}
inline double veclen(Point a) { return sqrt(dot(a, a));} struct Circle {
Point c;
double r;
Circle() {}
Circle(Point c, double r) : c(c), r(r) {}
Point point(double p) { return Point(c.x + r * cos(p), c.y + r * sin(p));}
void get() { scanf("%lf%lf%lf", &c.x, &c.y, &r);}
} ; bool ccint(Circle a, Circle b, double &a1, double &a2) { // ips for Circle-a (anti-clockwise p1->p2)
double d = veclen(a.c - b.c);
double r1 = a.r, r2 = b.r;
if (sgn(d - r1 - r2) >= ) return ;
if (sgn(fabs(r1 - r2) - d) >= ) return ;
double da = acos((sqr(d) + sqr(r1) - sqr(r2)) / ( * d * r1));
double ang = angle(b.c - a.c);
a1 = ang - da, a2 = ang + da;
return ;
} const int N = ;
Circle cir[N]; inline double gettri(Point o, Point a, Point b) { return cross(a - o, b - o) / ;}
inline double getarc(Circle o, double a, double b) { return (b - a) * sqr(o.r) / - gettri(o.c, o.point(a), o.point(b));} typedef pair<double, int> Event;
Event ev[N << ]; bool inside(int a, int n) {
double d, r1, r2;
for (int i = ; i < n; i++) {
if (i == a) continue;
d = veclen(cir[i].c - cir[a].c);
r1 = cir[i].r, r2 = cir[a].r;
if (sgn(r1 - r2) == && sgn(fabs(r1 - r2) - d) == ) {
if (i > a) return ;
continue;
}
if (sgn(r1 - r2) >= && sgn(fabs(r1 - r2) - d) >= ) return ;
}
return ;
} double getarea(int a, int n) {
if (inside(a, n)) return ;
Circle &c = cir[a];
double ret = ;
double a1, a2;
Point p1, p2;
int tt = ;
for (int i = ; i < n; i++) {
if (i == a) continue;
if (ccint(c, cir[i], a1, a2)) {
if (a2 > PI) ev[tt++] = Event(a1, ), ev[tt++] = Event(PI, -), ev[tt++] = Event(-PI, ), ev[tt++] = Event(a2 - * PI, -);
else if (a1 < -PI) ev[tt++] = Event(a1 + * PI, ), ev[tt++] = Event(PI, -), ev[tt++] = Event(-PI, ), ev[tt++] = Event(a2, -);
else ev[tt++] = Event(a1, ), ev[tt++] = Event(a2, -);
}
}
if (tt == ) return PI * sqr(c.r);
sort(ev, ev + tt);
int cnt = ;
double last = -PI;
for (int i = ; i < tt; i++) {
double cur = ev[i].x;
if (ev[i].y == && cnt == ) ret += getarc(c, last, cur) + cross(c.point(last), c.point(cur)) / ;
cnt += ev[i].y;
last = cur;
}
ret += getarc(c, last, PI) + cross(c.point(last), c.point(PI)) / ;
return ret;
} double work(int n) {
double ret = ;
for (int i = ; i < n; i++) ret += getarea(i, n);
return ret;
} int main() {
int n;
while (~scanf("%d", &n)) {
for (int i = ; i < n; i++) cir[i].get();
printf("%.3f\n", work(n));
}
return ;
}
——written by Lyon