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Viva Confetti
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 881 | Accepted: 361 |
Description
Do you know confetti? They are small discs of colored paper, and people throw them around during parties or festivals. Since people throw lots of confetti, they may end up stacked one on another, so there may be hidden ones underneath.
A handful of various sized confetti have been dropped on a table. Given their positions and sizes, can you tell us how many of them you can see?
The following figure represents the disc configuration for the first sample input, where the bottom disc is still visible.
Input
The input is composed of a number of configurations of the following form.
n
x1 y1 r1
x2 y2 r2
...
xn yn rn
The first line in a configuration is the number of discs in the configuration (a positive integer not more than 100), followed by one line descriptions of each disc : coordinates of its center and radius, expressed as real numbers in decimal notation, with up to 12 digits after the decimal point. The imprecision margin is +/- 5 x 10^(-13). That is, it is guaranteed that variations of less than +/- 5 x 10^(-13) on input values do not change which discs are visible. Coordinates of all points contained in discs are between -10 and 10.
Confetti are listed in their stacking order, x1 y1 r1 being the bottom one and xn yn rn the top one. You are observing from the top.
The end of the input is marked by a zero on a single line.
Output
For each configuration you should output the number of visible confetti on a single line.
Sample Input
3 0 0 0.5 -0.9 0 1.00000000001 0.9 0 1.00000000001 5 0 1 0.5 1 1 1.00000000001 0 2 1.00000000001 -1 1 1.00000000001 0 -0.00001 1.00000000001 5 0 1 0.5 1 1 1.00000000001 0 2 1.00000000001 -1 1 1.00000000001 0 0 1.00000000001 2 0 0 1.0000001 0 0 1 2 0 0 1 0.00000001 0 1 0
Sample Output
3 5 4 2 2
依照顺序摆放下n个圆，问最终有多少个圆是可见的。
想了好久，然后问了队友shu_mj，想了好一会才想通。
首先，可见部分的一部分的边界一定是圆弧。于是，我们可以先求出所有的圆相交划分的一小段一小段的圆弧，然后把这些小段圆弧的中点稍微往圆内移动一点以及往外移动一点。然后从后放的圆开始判断，最先出现在哪个圆中，那么这个圆就是可见的。
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 }
代码君