## poj1981 Circle and Points 单位圆覆盖问题

Circle and Points
 Time Limit: 5000MS Memory Limit: 30000K Total Submissions: 6850 Accepted: 2443 Case Time Limit: 2000MS

Description

You are given N points in the xy-plane. You have a circle of radius one and move it on the xy-plane, so as to enclose as many of the points as possible. Find how many points can be simultaneously enclosed at the maximum. A point is considered enclosed by a circle when it is inside or on the circle.

Fig 1. Circle and Points

Input

The input consists of a series of data sets, followed by a single line only containing a single character '0', which indicates the end of the input. Each data set begins with a line containing an integer N, which indicates the number of points in the data set. It is followed by N lines describing the coordinates of the points. Each of the N lines has two decimal fractions X and Y, describing the x- and y-coordinates of a point, respectively. They are given with five digits after the decimal point.

You may assume 1 <= N <= 300, 0.0 <= X <= 10.0, and 0.0 <= Y <= 10.0. No two points are closer than 0.0001. No two points in a data set are approximately at a distance of 2.0. More precisely, for any two points in a data set, the distance d between the two never satisfies 1.9999 <= d <= 2.0001. Finally, no three points in a data set are simultaneously very close to a single circle of radius one. More precisely, let P1, P2, and P3 be any three points in a data set, and d1, d2, and d3 the distances from an arbitrarily selected point in the xy-plane to each of them respectively. Then it never simultaneously holds that 0.9999 <= di <= 1.0001 (i = 1, 2, 3).

Output

For each data set, print a single line containing the maximum number of points in the data set that can be simultaneously enclosed by a circle of radius one. No other characters including leading and trailing spaces should be printed.

Sample Input

3
6.47634 7.69628
5.16828 4.79915
6.69533 6.20378
6
7.15296 4.08328
6.50827 2.69466
5.91219 3.86661
5.29853 4.16097
6.10838 3.46039
6.34060 2.41599
8
7.90650 4.01746
4.10998 4.18354
4.67289 4.01887
6.33885 4.28388
4.98106 3.82728
5.12379 5.16473
7.84664 4.67693
4.02776 3.87990
20
6.65128 5.47490
6.42743 6.26189
6.35864 4.61611
6.59020 4.54228
4.43967 5.70059
4.38226 5.70536
5.50755 6.18163
7.41971 6.13668
6.71936 3.04496
5.61832 4.23857
5.99424 4.29328
5.60961 4.32998
6.82242 5.79683
5.44693 3.82724
6.70906 3.65736
7.89087 5.68000
6.23300 4.59530
5.92401 4.92329
6.24168 3.81389
6.22671 3.62210
0

Sample Output

2
5
5
11

1 /**
2  * code generated by 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 //#####################
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 mp(X, Y) make_pair(X,Y)
38 #define rep(X, N) for(int X=0;X<N;X++)
39
40 //
41 // Created by xyiyy on 2015/8/10.
42 //
43
44 #ifndef JHELPER_EXAMPLE_PROJECT_P_HPP
45 #define JHELPER_EXAMPLE_PROJECT_P_HPP
46
47 const double EPS = 1e-9;
48
49 double add(double a, double b) {
50     if (fabs(a + b) < EPS * (fabs(a) + fabs(b)))return 0;
51     return a + b;
52 }
53
54 class P {
55 public:
56     double x, y;
57
58     P() { }
59
60     P(double x, double y) : x(x), y(y) { }
61
62     P  operator+(const P &p) {
64     }
65
66     P operator-(const P &p) {
68     }
69
70     P operator*(const double &d) {
71         return P(x * d, y * d);
72     }
73
74     P operator/(const double &d) {
75         return P(x / d, y / d);
76     }
77
78     double dot(P p) {
79         return add(x * p.x, y * p.y);
80     }
81
82
83     double abs() {
84         return sqrt(abs2());
85     }
86
87     double abs2() {
88         return dot(*this);
89     }
90
91 };
92
93
94
95 //直线和直线的交点
96 /*P isLL(P p1,P p2,P q1,P q2){
97     double d = (q2 - q1).det(p2 - p1);
98     if(sig(d)==0)return NULL;
99     return intersection(p1,p2,q1,q2);
100 }*/
101
102
103 //四点共圆判定
104 /*bool onC(P p1,P p2,P p3,P p4){
105     P c = CCenter(p1,p2,p3);
106     if(c == NULL) return false;
107     return add((c - p1).abs2(), -(c - p4).abs2()) == 0;
108 }*/
109
110 //三点共圆的圆心
111
112
113 #endif //JHELPER_EXAMPLE_PROJECT_P_HPP
114
115 const int MAXN = 710;
116 P ps[MAXN];
117 pair<double, bool> arc[MAXN];
118
119 class poj1981 {
120 public:
121     void solve(std::istream &in, std::ostream &out) {
122         int n;
123         P t;
124         while (in >> n && n) {
125             rep(i, n)in >> ps[i].x >> ps[i].y;
126             int ans = 1;
127             rep(i, n) {
128                 int num = 0;
129                 rep(j, n) {
130                     if (i == j)continue;
131                     double d;
132                     if ((d = (ps[i] - ps[j]).abs()) <= 2) {
133                         double a = acos(d / 2);
134                         double b = atan2((ps[j].y - ps[i].y), (ps[j].x - ps[i].x));
135                         arc[num++] = mp(b - a, 1);
136                         arc[num++] = mp(b + a, 0);
137                     }
138                 }
139                 sort(arc, arc + num);
140                 int res = 1;
141                 rep(j, num) {
142                     if (arc[j].second)res++;
143                     else res--;
144                     ans = max(ans, res);
145                 }
146             }
147             out << ans << endl;
148         }
149     }
150 };
151
152 int main() {
153     std::ios::sync_with_stdio(false);
154     std::cin.tie(0);
155     poj1981 solver;
156     std::istream &in(std::cin);
157     std::ostream &out(std::cout);
158     solver.solve(in, out);
159     return 0;
160 }

posted on 2015-08-31 15:46  xyiyy  阅读(...)  评论(...编辑  收藏

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