ural 1333 化平面为点计算覆盖率

题目链接：http://acm.timus.ru/problem.aspx?space=1&num=1333

#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() {}
     Circle(Point c,double r): c(c),r(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; }

Point read_point(){
    Point A;
    scanf("%lf %lf",&A.x,&A.y);
    return A;
}

/*************************************分 割 线*****************************************/
const int maxn = 15;
Circle C[maxn];

int main()
{
    //freopen("E:\acm\input.txt","r",stdin);

    int N;
    cin>>N;
    for(int i=1;i<=N;i++){
        scanf("%lf %lf %lf",&C[i].c.x,&C[i].c.y,&C[i].r);
    }

    double cnt = 0;
    for(double x = 0.000;x<=1.000;x+=0.001)
        for(double y = 0.000;y<=1.000;y+=0.001){
            Point A = Point(x,y);
            for(int i=1;i<=N;i++){
                if(dcmp(Length(A-C[i].c)-C[i].r) < 0){
                    cnt ++;
                    break;
                }
            }
    }
    double prec = cnt/1e6;

    cout<<prec*100<<endl;
}