学步园

D.Mutikill

给定圆半径长，求最多可覆盖平面上多少点。

定长圆覆盖问题，和poj 1981可说是一模一样。

然后僵尸可能一个都木有（所以这题一定是搞计算几何的人来坑搞计算几何的了，或者是我太碍板了= =。。）

#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <stdio.h>
#include <math.h>
using namespace std;
#define LL long long
#define MAXN 310
#define eps 1e-10

const double pi = acos(-1.0);

int sig(double x)
{
    return (x > eps) - (x < -eps);
}

typedef struct Point
{
        double x, y;
        Point() {}
        Point(double _x, double _y):
                x(_x), y(_y) {}
        bool operator <(const Point &argu) const
        {
                if(sig(x - argu.x) == 0) return y < argu.y;
                return x < argu.x;
        }
        double dis(const Point &argu) const
        {
            return sqrt((x - argu.x) * (x - argu.x) + (y - argu.y) * (y - argu.y));
        }
        double dis2(const Point &argu) const
        {
            return (x - argu.x) * (x - argu.x) + (y - argu.y) * (y - argu.y);
        }
        double operator ^(const Point &argu) const
        {
            return x * argu.y - y * argu.x;
        }
        double operator *(const Point &argu) const
        {
            return x * argu.x + y * argu.y;
        }
        Point operator -(const Point &argu) const
        {
            return Point(x - argu.x, y - argu.y);
        }
        double len2() const
        {
            return x * x + y * y;
        }
        double len() const
        {
            return sqrt(x * x + y * y);
        }
        void in()
        {
            scanf("%lf%lf", &x, &y);
        }
        void out()
        {
            printf("%.10lf %.10lf\n", x, y);
        }
}Vector;

struct Arc
{
    double ang;
    int in;
    bool operator <(const Arc &argu) const
    {
        return sig(ang - argu.ang) ? (ang < argu.ang) : (in > argu.in);
    }
}arc[MAXN];

int n;
double r;
Point p[MAXN];

int max(int a, int b) { return a > b ? a : b; }

int MAX_CirCle_Cover(double r)
{
    int ret = 0;
    for(int i = 0; i < n; i++)
    {
        int cnt = 0;
        for(int j = 0; j < n; j++)
        {
            if(i == j) continue;
            double d = p[i].dis(p[j]);
            if(sig(d - 2.0 * r) > 0) continue;
            double theta = atan2(p[i].y - p[j].y, p[i].x - p[j].x);
            if(theta < 0) theta += 2 * pi;
            double phi = acos(d / (2.0 * r));
            arc[cnt].ang = theta - phi + 2 * pi; arc[cnt++].in = 1;
            arc[cnt].ang = theta + phi + 2 * pi; arc[cnt++].in = 0;
        }
        sort(arc, arc + cnt);
        int tmp = 0;
        for(int i = 0; i < cnt; i++)
        {
            if(arc[i].in) tmp++;
            else tmp--;
            ret = max(ret, tmp);
        }
    }
    return ret + 1;
}

int work()
{
    scanf("%lf%d", &r, &n);
    if(sig(r) == 0 || n == 0) return 0;
    for(int i = 0; i < n; i++) p[i].in();
    return MAX_CirCle_Cover(r);
}

int main()
{
//    freopen("D_std.in", "r", stdin);

    int c; scanf("%d", &c);
    while(c--)
    {
        printf("%d\n", work());
    }
    return 0;
}