Catching Fish
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1827 Accepted Submission(s): 725
Problem Description
Ignatius
likes catching fish very much. He has a fishnet whose shape is a circle
of radius one. Now he is about to use his fishnet to catch fish. All
the fish are in the lake, and we assume all the fish will not move when
Ignatius catching them. Now Ignatius wants to know how many fish he can
catch by using his fishnet once. We assume that the fish can be regard
as a point. So now the problem is how many points can be enclosed by a
circle of radius one.
Note: If a fish is just on the border of the fishnet, it is also caught by Ignatius.
Input
The
input contains several test cases. The first line of the input is a
single integer T which is the number of test cases. T test cases follow.
Each
test case starts with a positive integer N(1<=N<=300) which
indicate the number of fish in the lake. Then N lines follow. Each line
contains two floating-point number X and Y (0.0<=X,Y<=10.0). You
may assume no two fish will at the same point, and no two fish are
closer than 0.0001, no two fish in a test case are approximately at a
distance of 2.0. In other words, if the distance between the fish and
the centre of the fishnet is smaller 1.0001, we say the fish is also
caught.
Output
For each test case, you should output the maximum number of fish Ignatius can catch by using his fishnet once.
Sample Input
4
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
Sample Output
2
5
5
11
题意：
给你N个点，计算出单位圆内最多包含多少个点
思路：
每两个点每举出一个圆
AC代码：
1 # include <bits/stdc++.h>
2 using namespace std;
3 const double eps = 1e-6;
4
5 struct Point
6 {
7 double x;
8 double y;
9 }p[305];
10
11 double dis2(Point a, Point b)
12 {
13 return (b.y - a.y) * (b.y - a.y) + (b.x - a.x) * (b.x - a.x);
14 }
15
16 Point center(Point a, Point b)
17 {
18 Point aa, bb, mid;
19 aa.x = b.x - a.x;
20 aa.y = b.y - a.y;
21 mid.x = (a.x + b.x) / 2;
22 mid.y = (a.y + b.y) / 2;
23 double c = sqrt(1.0 - dis2(a, mid));
24 if(fabs(aa.y) < eps)
25 {
26 bb.x = mid.x;
27 bb.y = mid.y + c;
28 }
29 else
30 {
31 double ang = atan(-aa.x / aa.y);
32 bb.x = mid.x + c * cos(ang);
33 bb.y = mid.y + c * sin(ang);
34 }
35 return bb;
36 }
37
38 int main()
39 {
40 int T;
41 scanf("%d", &T);
42 while(T--)
43 {
44 int n;
45 scanf("%d", &n);
46 for(int i = 0; i < n; i++)
47 {
48 scanf("%lf%lf", &p[i].x, &p[i].y);
49 }
50 Point cen;
51 int ans = 1, tep;
52 for(int i = 0; i < n; i++)
53 {
54 for(int j = i + 1; j < n; j++)
55 {
56 if(dis2(p[i], p[j]) > 4)
57 continue;
58 cen = center(p[i], p[j]);
59 tep = 0;
60 for(int k = 0; k < n; k++)
61 {
62 if(dis2(p[k], cen) <= 1 + eps)
63 tep++;
64 }
65 if(ans < tep)
66 ans = tep;
67 }
68 }
69 printf("%d\n", ans);
70 }
71
72
73 return 0;
74 }