HDU 1077 Catching Fish（用单位圆尽可能围住多的点）

Posted 2020-12-05 yinbiao

传送门：http://acm.hdu.edu.cn/showproblem.php?pid=1077

Catching Fish

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2636    Accepted Submission(s): 969

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

 

题目意思：

告诉你一些点，要求你用单位圆尽可能围住多的点，问你最多围住多少点？

在单位圆边缘的点也算围住

做法：

每次枚举两个距离小于2.001的点在单位圆上（距离大于2.001的点不可能在同一个单位圆上）

根据这两个点的位置，可以确定单位圆圆心的位置，再来计算哪些点到圆心的距离小于等于1.001（就是围住了哪些点）

然后在每次枚举得到的围住的点中找到最大值

就是最多围住的点数！

注意两个点可以确定两个单位圆，有两个圆心，所以需要算两次

 

code：

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
#define max_v 305
double p[max_v][2];
int n;
double dis(double x1,double y1,double x2,double y2)//两点距离
{
    return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}
double f(int i,int j)
{
    double x1,y1,x2,y2,x3,y3,x4,y4,x5,y5;
    x1=p[i][0];
    y1=p[i][1];

    x2=p[j][0];
    y2=p[j][1];

    double s=dis(x1,y1,x2,y2);
    double xx=(y2-y1)/s;//(xx,yy)相当于与弦长垂直的单位法向量
    double yy=(x1-x2)/s;

    s=s/2.0;
    s=sqrt(1.0-s*s);//圆心与两点弦长的距离

    x3=(x1+x2)/2.0;
    y3=(y1+y2)/2.0;//(x3,y3)是（x1,y1),(x2,y2)的中点

    int c1=0,c2=0;

    x4=x3+s*xx;
    y4=y3+s*yy;//(x4,y4)现在是圆心
    for(int i=0;i<n;i++)
    {
        if(dis(x4,y4,p[i][0],p[i][1])<1.0001)
            c1++;
    }

    x5=x3-s*xx;
    y5=y3-s*yy;//(x5,y5)现在是圆心
    for(int i=0;i<n;i++)
    {
        if(dis(x5,y5,p[i][0],p[i][1])<1.0001)
            c2++;
    }

    if(c1>c2)
        return c1;
    else
        return c2;

}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i=0;i<n;i++)
        {
            scanf("%lf %lf",&p[i][0],&p[i][1]);
        }
        int temp,sum=1;
        
        //暴力，每次让两点恰好位于单位圆上，算出圆心。然后找覆盖点的数目，输出最大的
        for(int i=0;i<n;i++)
        {
            for(int j=i+1;j<n;j++)
            {
                if(dis(p[i][0],p[i][1],p[j][0],p[j][1])<2.0001)//距离大于2.0001的点可以不枚举，因为这两点肯定不在一个单位圆上
                {
                    temp=f(i,j);
                    if(sum<temp)
                    {
                        sum=temp;
                    }

                }
            }
        }
        printf("%d
",sum);
    }
    return 0;
}