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枚举一下两圆，处理出所有弧，然后判断每段弧的中点是否可见，可见的话该弧所在的圆也可见，以及该段弧下面的圆也可见。
#include <cstdio>
#include <cstring>
#include <cmath>
#include <vector>
#include <algorithm>
using namespace std;
const double pi = 4 * atan(1);
const double eps = 1e-14;
inline int dcmp (double x) { if (fabs(x) < eps) return 0; else return x < 0 ? -1 : 1; }
inline double getDistance (double x, double y) { return sqrt(x * x + y * y); }
struct Point {
double x, y;
Point (double x = 0, double y = 0): x(x), y(y) {}
void read () { scanf("%lf%lf", &x, &y); }
void write () { printf("%lf %lf", x, y); }
bool operator == (const Point& u) const { return dcmp(x - u.x) == 0 && dcmp(y - u.y) == 0; }
bool operator != (const Point& u) const { return !(*this == u); }
bool operator < (const Point& u) const { return x < u.x || (x == u.x && y < u.y); }
bool operator > (const Point& u) const { return u < *this; }
bool operator <= (const Point& u) const { return *this < u || *this == u; }
bool operator >= (const Point& u) const { return *this > u || *this == u; }
Point operator + (const Point& u) { return Point(x + u.x, y + u.y); }
Point operator - (const Point& u) { return Point(x - u.x, y - u.y); }
Point operator * (const double u) { return Point(x * u, y * u); }
Point operator / (const double u) { return Point(x / u, y / u); }
};
typedef Point Vector;
struct Line {
double a, b, c;
Line (double a = 0, double b = 0, double c = 0): a(a), b(b), c(c) {}
};
struct Circle {
Point o;
double r;
Circle () {}
Circle (Point o, double r = 0): o(o), r(r) {}
Point point(double rad) { return Point(o.x + cos(rad)*r, o.y + sin(rad)*r); }
void read() { scanf("%lf%lf%lf", &o.x, &o.y, &r); }
};
namespace Punctual {
double getDistance (Point a, Point b) { double x=a.x-b.x, y=a.y-b.y; return sqrt(x*x + y*y); }
};
namespace Vectorial {
/* 点积: 两向量长度的乘积再乘上它们夹角的余弦, 夹角大于90度时点积为负 */
double getDot (Vector a, Vector b) { return a.x * b.x + a.y * b.y; }
/* 叉积: 叉积等于两向量组成的三角形有向面积的两倍, cross(v, w) = -cross(w, v) */
double getCross (Vector a, Vector b) { return a.x * b.y - a.y * b.x; }
double getLength (Vector a) { return sqrt(getDot(a, a)); }
double getAngle (Vector u) { return atan2(u.y, u.x); }
double getAngle (Vector a, Vector b) { return acos(getDot(a, b) / getLength(a) / getLength(b)); }
Vector rotate (Vector a, double rad) { return Vector(a.x*cos(rad)-a.y*sin(rad), a.x*sin(rad)+a.y*cos(rad)); }
/* 单位法线 */
Vector getNormal (Vector a) { double l = getLength(a); return Vector(-a.y/l, a.x/l); }
};
namespace Linear {
using namespace Vectorial;
Line getLine (double x1, double y1, double x2, double y2) { return Line(y2-y1, x1-x2, y1*(x2-x1)-x1*(y2-y1)); }
Line getLine (double a, double b, Point u) { return Line(a, -b, u.y * b - u.x * a); }
bool getIntersection (Line p, Line q, Point& o) {
if (fabs(p.a * q.b - q.a * p.b) < eps)
return false;
o.x = (q.c * p.b - p.c * q.b) / (p.a * q.b - q.a * p.b);
o.y = (q.c * p.a - p.c * q.a) / (p.b * q.a - q.b * p.a);
return true;
}
/* 直线pv和直线qw的交点 */
bool getIntersection (Point p, Vector v, Point q, Vector w, Point& o) {
if (dcmp(getCross(v, w)) == 0) return false;
Vector u = p - q;
double k = getCross(w, u) / getCross(v, w);
o = p + v * k;
return true;
}
/* 点p到直线ab的距离 */
double getDistanceToLine (Point p, Point a, Point b) { return fabs(getCross(b-a, p-a) / getLength(b-a)); }
double getDistanceToSegment (Point p, Point a, Point b) {
if (a == b) return getLength(p-a);
Vector v1 = b - a, v2 = p - a, v3 = p - b;
if (dcmp(getDot(v1, v2)) < 0) return getLength(v2);
else if (dcmp(getDot(v1, v3)) > 0) return getLength(v3);
else return fabs(getCross(v1, v2) / getLength(v1));
}
/* 点p在直线ab上的投影 */
Point getPointToLine (Point p, Point a, Point b) { Vector v = b-a; return a+v*(getDot(v, p-a) / getDot(v,v)); }
/* 判断线段是否存在交点 */
bool haveIntersection (Point a1, Point a2, Point b1, Point b2) {
double c1=getCross(a2-a1, b1-a1), c2=getCross(a2-a1, b2-a1), c3=getCross(b2-b1, a1-b1), c4=getCross(b2-b1,a2-b1);
return dcmp(c1)*dcmp(c2) < 0 && dcmp(c3)*dcmp(c4) < 0;
}
/* 判断点是否在线段上 */
bool onSegment (Point p, Point a, Point b) { return dcmp(getCross(a-p, b-p)) == 0 && dcmp(getDot(a-p, b-p)) < 0; }
}
namespace Triangular {
using namespace Vectorial;
double getAngle (double a, double b, double c) { return acos((a*a+b*b-c*c) / (2*a*b)); }
double getArea (double a, double b, double c) { double s =(a+b+c)/2; return sqrt(s*(s-a)*(s-b)*(s-c)); }
double getArea (double a, double h) { return a * h / 2; }
double getArea (Point a, Point b, Point c) { return fabs(getCross(b - a, c - a)) / 2; }
};
namespace Polygonal {
using namespace Vectorial;
double getArea (Point* p, int n) {
double ret = 0;
for (int i = 1; i < n-1; i++)
ret += getCross(p[i]-p[0], p[i+1]-p[0]);
return fabs(ret)/2;
}
};
namespace Circular {
using namespace Vectorial;
using namespace Linear;
/* 直线和原的交点 */
int getLineCircleIntersection (Point p, Point q, Circle O, double& t1, double& t2, vector<Point>& sol) {
Vector v = q - p;
/* 使用前需清空sol */
//sol.clear();
double a = v.x, b = p.x - O.o.x, c = v.y, d = p.y - O.o.y;
double e = a*a+c*c, f = 2*(a*b+c*d), g = b*b+d*d-O.r*O.r;
double delta = f*f - 4*e*g;
if (dcmp(delta) < 0) return 0;
if (dcmp(delta) == 0) {
t1 = t2 = -f / (2 * e);
sol.push_back(p + v * t1);
return 1;
}
t1 = (-f - sqrt(delta)) / (2 * e); sol.push_back(p + v * t1);
t2 = (-f + sqrt(delta)) / (2 * e); sol.push_back(p + v * t2);
return 2;
}
/* 圆和圆的交点 */
int getCircleCircleIntersection (Circle o1, Circle o2, vector<Point>& sol) {
double d = getLength(o1.o - o2.o);
if (dcmp(d) == 0) {
if (dcmp(o1.r - o2.r) == 0) return -1;
return 0;
}
if (dcmp(o1.r + o2.r - d) < 0) return 0;
if (dcmp(fabs(o1.r-o2.r) - d) > 0) return 0;
double a = getAngle(o2.o - o1.o);
double da = acos((o1.r*o1.r + d*d - o2.r*o2.r) / (2*o1.r*d));
Point p1 = o1.point(a-da), p2 = o1.point(a+da);
sol.push_back(p1);
if (p1 == p2) return 1;
sol.push_back(p2);
return 2;
}
/* 过定点作圆的切线 */
int getTangents (Point p, Circle o, Vector* v) {
Vector u = o.o - p;
double d = getLength(u);
if (d < o.r) return 0;
else if (dcmp(d - o.r) == 0) {
v[0] = rotate(u, pi / 2);
return 1;
} else {
double ang = asin(o.r / d);
v[0] = rotate(u, -ang);
v[1] = rotate(u, ang);
return 2;
}
}
/* a[i] 和 b[i] 分别是第i条切线在O1和O2上的切点 */
int getTangents (Circle o1, Circle o2, Point* a, Point* b) {
int cnt = 0;
if (o1.r < o2.r) { swap(o1, o2); swap(a, b); }
double d2 = getLength(o1.o - o2.o); d2 = d2 * d2;
double rdif = o1.r - o2.r, rsum = o1.r + o2.r;
if (d2 < rdif * rdif) return 0;
if (dcmp(d2) == 0 && dcmp(o1.r - o2.r) == 0) return -1;
double base = getAngle(o2.o - o1.o);
if (dcmp(d2 - rdif * rdif) == 0) {
a[cnt] = o1.point(base); b[cnt] = o2.point(base); cnt++;
return cnt;
}
double ang = acos( (o1.r - o2.r) / sqrt(d2) );
a[cnt] = o1.point(base+ang); b[cnt] = o2.point(base+ang); cnt++;
a[cnt] = o1.point(base-ang); b[cnt] = o2.point(base-ang); cnt++;
if (dcmp(d2 - rsum * rsum) == 0) {
a[cnt] = o1.point(base); b[cnt] = o2.point(base); cnt++;
} else if (d2 > rsum * rsum) {
double ang = acos( (o1.r + o2.r) / sqrt(d2) );
a[cnt] = o1.point(base+ang); b[cnt] = o2.point(base+ang); cnt++;
a[cnt] = o1.point(base-ang); b[cnt] = o2.point(base-ang); cnt++;
}
return cnt;
}
/* 三点确定外切圆 */
Circle CircumscribedCircle(Point p1, Point p2, Point p3) {
double Bx = p2.x - p1.x, By = p2.y - p1.y;
double Cx = p3.x - p1.x, Cy = p3.y - p1.y;
double D = 2 * (Bx * Cy - By * Cx);
double cx = (Cy * (Bx * Bx + By * By) - By * (Cx * Cx + Cy * Cy)) / D + p1.x;
double cy = (Bx * (Cx * Cx + Cy * Cy) - Cx * (Bx * Bx + By * By)) / D + p1.y;
Point p = Point(cx, cy);
return Circle(p, getLength(p1 - p));
}
/* 三点确定内切圆 */
Circle InscribedCircle(Point p1, Point p2, Point p3) {
double a = getLength(p2 - p3);
double b = getLength(p3 - p1);
double c = getLength(p1 - p2);
Point p = (p1 * a + p2 * b + p3 * c) / (a + b + c);
return Circle(p, getDistanceToLine(p, p1, p2));
}
};
using namespace Vectorial;
using namespace Circular;
const int maxn = 105;
int N, vis[maxn];
Circle o[maxn];
vector<double> rad[maxn];
int main () {
while (scanf("%d", &N) == 1 && N) {
memset(vis, 0, sizeof(vis));
for (int i = 1; i <= N; i++) o[i].read();
for (int i = 1; i <= N; i++) {
vector<Point> sol;
for (int j = 1; j <= N; j++)
getCircleCircleIntersection (o[i], o[j], sol);
rad[i].clear();
rad[i].push_back(0);
for (int j = 0; j < sol.size(); j++)
rad[i].push_back(getAngle(sol[j] - o[i].o));
if (rad[i].size() == 0) rad[i].push_back(2*pi);
sort(rad[i].begin(), rad[i].end());
}
for (int i = 1; i <= N; i++) {
for (int j = 0; j < rad[i].size(); j++) {
Point u;
if (j)
u = o[i].point((rad[i][j] + rad[i][j-1]) / 2);
else
u = o[i].point((rad[i][j] + rad[i][rad[i].size()-1]) / 2 + pi);
bool flag = true;
for (int k = i + 1; k <= N; k++) {
if (dcmp(getLength(u - o[k].o) - o[k].r) <= 0) {
flag = false; break;
}
}
if (flag) {
vis[i] = 1;
for (int k = i-1; k >= 0; k--) {
if (dcmp(getLength(u - o[k].o) - o[k].r) <= 0) {
vis[k] = 1; break;
}
}
}
}
}
int ans = 0;
for (int i = 1; i <= N; i++)
if (vis[i]) ans++;
printf("%d\n", ans);
}
return 0;
}
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原文地址：http://blog.csdn.net/keshuai19940722/article/details/47731713
