poj1418 Viva Confetti 判断圆是否可见

  1 /**
  2  * code generated by JHelper
  3  * More info: https://github.com/AlexeyDmitriev/JHelper
  4  * @author xyiyy @https://github.com/xyiyy
  5  */
  6 
  7 #include <iostream>
  8 #include <fstream>
  9 
 10 //#####################
 11 //Author:fraud
 12 //Blog: http://www.cnblogs.com/fraud/
 13 //#####################
 14 //#pragma comment(linker, "/STACK:102400000,102400000")
 15 #include <iostream>
 16 #include <sstream>
 17 #include <ios>
 18 #include <iomanip>
 19 #include <functional>
 20 #include <algorithm>
 21 #include <vector>
 22 #include <string>
 23 #include <list>
 24 #include <queue>
 25 #include <deque>
 26 #include <stack>
 27 #include <set>
 28 #include <map>
 29 #include <cstdio>
 30 #include <cstdlib>
 31 #include <cmath>
 32 #include <cstring>
 33 #include <climits>
 34 #include <cctype>
 35 
 36 using namespace std;
 37 #define pb(X) push_back(X)
 38 #define rep(X, N) for(int X=0;X<N;X++)
 39 #define ALL(X) (X).begin(),(X).end()
 40 
 41 //
 42 // Created by xyiyy on 2015/8/10.
 43 //
 44 
 45 #ifndef JHELPER_EXAMPLE_PROJECT_P_HPP
 46 #define JHELPER_EXAMPLE_PROJECT_P_HPP
 47 
 48 const double EPS = 4e-13;
 49 
 50 double add(double a, double b) {
 51     if (fabs(a + b) < EPS * (fabs(a) + fabs(b)))return 0;
 52     return a + b;
 53 }
 54 
 55 class P {
 56 public:
 57     double x, y;
 58 
 59     P() { }
 60 
 61     P(double x, double y) : x(x), y(y) { }
 62 
 63     P  operator+(const P &p) {
 64         return P(add(x, p.x), add(y, p.y));
 65     }
 66 
 67     P operator-(const P &p) {
 68         return P(add(x, -p.x), add(y, -p.y));
 69     }
 70 
 71     P operator*(const double &d) {
 72         return P(x * d, y * d);
 73     }
 74 
 75     P operator/(const double &d) {
 76         return P(x / d, y / d);
 77     }
 78 
 79     double dot(P p) {
 80         return add(x * p.x, y * p.y);
 81     }
 82 
 83 
 84     double abs() {
 85         return sqrt(abs2());
 86     }
 87 
 88     double abs2() {
 89         return dot(*this);
 90     }
 91 
 92 };
 93 
 94 
 95 
 96 //求两圆的极角 以p为中心
 97 double polarangle(P p, P q) {
 98     return atan2(q.y - p.y, q.x - p.x);
 99 }
100 
101 
102 #endif //JHELPER_EXAMPLE_PROJECT_P_HPP
103 
104 const long double PI2 = 2 * acos(-1.0);
105 
106 long double update(double x) {
107     while (x < 0.0)x += PI2;
108     while (x >= PI2)x -= PI2;
109     return x;
110 }
111 
112 class poj1418 {
113 public:
114     void solve(std::istream &in, std::ostream &out) {
115         int n;
116         P t;
117         while (in >> n && n) {
118             vector<P> ps;
119             vector<double> rs;
120             vector<bool> cansee(n, 0);
121             rep(i, n) {
122                 double x, y, r;
123                 in >> x >> y >> r;
124                 ps.pb(P(x, y));
125                 rs.pb(r);
126             }
127             rep(i, n) {
128                 vector<double> pp;
129                 pp.pb(0.0);
130                 pp.pb(PI2);
131                 rep(j, n) {
132                     double a = rs[i];
133                     double d = (ps[i] - ps[j]).abs();
134                     double b = rs[j];
135                     if (a + b < d || a + d < b || b + d < a)continue;
136                     double theta = acos((a * a + d * d - b * b) / (2 * a * d));
137                     double alpha = polarangle(ps[i], ps[j]);
138                     pp.pb(update(alpha - theta));
139                     pp.pb(update(alpha + theta));
140                 }
141                 sort(ALL(pp));
142                 rep(j, pp.size() - 1) {
143                     double theta = (pp[j] + pp[j + 1]) / 2;
144                     for (int k = -1; k <= 1; k += 2) {
145                         t.x = ps[i].x + (rs[i] + k * EPS) * cos(theta);
146                         t.y = ps[i].y + (rs[i] + k * EPS) * sin(theta);
147                         int gao = n - 1;
148                         for (; gao >= 0; gao--) {
149                             if ((ps[gao] - t).abs() < rs[gao])break;
150                         }
151                         if (gao != -1)cansee[gao] = 1;
152                     }
153                 }
154             }
155             out << count(ALL(cansee), 1) << endl;
156         }
157     }
158 };
159 
160 int main() {
161     std::ios::sync_with_stdio(false);
162     std::cin.tie(0);
163     poj1418 solver;
164     std::istream &in(std::cin);
165     std::ostream &out(std::cout);
166     solver.solve(in, out);
167     return 0;
168 }