Ural 1332 把圆细分+圆内切，内含关系判定

题目链接：http://acm.timus.ru/problem.aspx?space=1&num=1332

#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;

const double eps = 1e-8;
const double PI = acos(-1.0);
const double INF = 1000000000000000.000;

struct Point{
    double x,y;
    Point(double x=0, double y=0) : x(x),y(y){ }    //构造函数
};
typedef Point Vector;

struct Circle{
     Point C;
     double r;

     Circle(Point C=Point(0,0),double r=0.0): C(C),r(r) {}
     Point getpoint(double ang){   //可以用来求圆上与x正半轴成ang度的点的坐标。
        return Point(C.x +cos(ang)*r,C.y + sin(ang)*r);
     }
};

Vector operator + (Vector A , Vector B){return Vector(A.x+B.x,A.y+B.y);}
Vector operator - (Vector A , Vector B){return Vector(A.x-B.x,A.y-B.y);}
Vector operator * (double p,Vector A){return Vector(A.x*p,A.y*p);}
Vector operator / (Vector A , double p){return Vector(A.x/p,A.y/p);}

bool operator < (const Point& a,const Point& b){
    return a.x < b.x ||( a.x == b.x && a.y < b.y);
}

int dcmp(double x){
    if(fabs(x) < eps) return 0;
    else              return x < 0 ? -1 : 1;
}
bool operator == (const Point& a, const Point& b){
    return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;
}

///向量（x,y）的极角用atan2(y,x);
inline double Dot(Vector A, Vector B){ return A.x*B.x + A.y*B.y; }
inline double Length(Vector A)    { return sqrt(Dot(A,A)); }
inline double Angle(Vector A, Vector B)  { return acos(Dot(A,B) / Length(A) / Length(B)); }
double Cross(Vector A, Vector B)  { return A.x*B.y - A.y * B.x; }


Vector vecunit(Vector v){ return v / Length(v);} //单位向量


/*************************************分 割 线*****************************************/


const int maxn = 105;

Circle cir[maxn];
int N;
double R,r;

int main()
{
    //freopen("E:\acm\input.txt","r",stdin);
    cin>>N;
    for(int i=1;i<=N;i++){
       scanf("%lf %lf",&cir[i].C.x,&cir[i].C.y);
    }
    scanf("%lf %lf",&R,&r);

    if(dcmp(R-r) < 0){
        printf("0
");
        return 0;
    }

    Circle NewC;
    NewC.r = R;

    for(int i=1;i<=N;i++){
        cir[i].r = r;
    }
    double ave = 2*PI / 3000;
    int ans = 0;
    for(int i=1;i<=N;i++){
        for(double ang=0;ang<=2*PI;ang+=ave){
            int temp = 1;
            Point P = cir[i].getpoint(ang);   //在圆上取点。

            NewC.C = P + R * (cir[i].C-P)/Length(cir[i].C-P);

            for(int j=1;j<=N;j++){
                if(i == j) continue;
                if(dcmp(fabs(NewC.r-cir[j].r) - Length(NewC.C - cir[j].C)) >= 0) {
                     temp++;
                }
            }
            ans = max(ans,temp);
        }
    }

    printf("%d
",ans);
}