2015 HNU Warm Up 06

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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-10const 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;}

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