Catching Fish[HDU1077]

最新推荐文章于 2024-07-30 10:30:00 发布

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最新推荐文章于 2024-07-30 10:30:00 发布

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CC 4.0 BY-SA版权

文章标签： java

原文链接：http://www.cnblogs.com/dramstadt/p/3200204.html

本文介绍了一个基于几何计算的算法，该算法用于解决在一个单位圆内能捕获的最大鱼的数量问题。通过计算点之间的距离和利用几何性质确定圆心位置来找出最多能包含多少个点（鱼）的解决方案。

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Catching Fish

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1052 Accepted Submission(s): 385

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

Author
Ignatius.L

#include <cmath>   
#include <cstdio>   
  
using namespace std;  
  
const static double  eps = 1e-6;  
  
struct Point  
{  
    double x,y;  
};  
double distancess(Point a,Point b)  
{  
    return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);  
}  
Point look_center(Point a,Point b)  
{  
    Point aa,bb,mid;  
    aa.x = b.x-a.x;  
    aa.y = b.y-a.y;  
    mid.x = (a.x+b.x)/2.0;  
    mid.y = (a.y+b.y)/2.0;  
    double dist = distancess(a,mid);  
    double c = sqrt(1.0-dist);  
    if(fabs(aa.y)<eps)  
    {  
        bb.x = mid.x;  
        bb.y = mid.y+c;  
    }  
      
    else  
    {  
        double ang = atan(-aa.x/aa.y);  
        bb.x = mid.x + c*cos(ang);  
        bb.y = mid.y + c*sin(ang);  
    }  
    return bb;  
}  
int main()  
{  
    int test;  
    Point p[305],a,b,c;  
    int n;  
    scanf("%d",&test);  
    while(test--)  
    {  
        scanf("%d",&n);  
        for(int i=0; i<n;i++)  
            scanf("%lf%lf",&p[i].x,&p[i].y);  
  
        int ans = 1;  
        int temp = 0;  
        for(int i=0; i<n; i++)  
            for(int j=i+1;j<n;j++)  
            {     
                  
                if(distancess(p[i],p[j])>4) continue;  
                a = look_center(p[i],p[j]);  
                temp = 0;  
                for(int k=0;k<n;k++)  
                {  
                    if(distancess(a,p[k])<=1.000001) temp++;  
                }  
                if(ans<temp) ans = temp;  
            }  
            printf("%d\n",ans);  
    }  
    return 0;  
  
}