zoukankan      html  css  js  c++  java
  • HDU 1077

    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
    ", ans);
    70     }
    71     
    72     
    73     return 0;
    74 }
    View Code
     
    生命不息,奋斗不止,这才叫青春,青春就是拥有热情相信未来。
  • 相关阅读:
    好吧,左小波出山了——ie8兼容indexOf问题
    jmeter负载机运行/添加压力机/分布式
    jmeter操作数据库
    Charles手机抓包设置&无法打开火狐网页设置
    python学习-Day1-接口测试
    动态SQL
    MyBatis缓存
    正则表达式
    MyBatis配置文件的配置说明
    几种数据源的配置
  • 原文地址:https://www.cnblogs.com/lyf-acm/p/5827269.html
Copyright © 2011-2022 走看看