[SPOJ]CIRU 圆并

存代码

新的编程方法？ 基本过程依赖于对象？

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <cmath>
using namespace std;
const int N = 1100;
const double eps = 1e-8, PI = 3.14159265358979;
int n;bool v[N];
inline double sqr (double x) {return x * x; }
inline int dcmp (double x) { return x < -eps ? -1 : x > eps; }
struct vec {
    double x, y;
    vec () {}
    vec (double _x, double _y) : x (_x), y (_y) {}

    double dot (const vec & a) { return x * a.x + y * a.y; }
    double cross (const vec & a) { return x * a.y - y * a.x; }
    double len () { return sqrt (x * x + y * y); }
    vec pull (double t) {
        return vec (x * t, y * t);
    }
    vec rot (double cos, double sin) {
        //printf ("%lf %lf\n", x * cos + y * sin, x * -sin + y * cos);
        return vec (x * cos + y * sin, x * -sin + y * cos); }
};
typedef vec point;
inline vec operator - (const point & a, const point & b)
{
    return vec (a.x - b.x, a.y - b.y);
}
inline point operator + (const point & a, const vec & b)
{
    return point (a.x + b.x, a.y + b.y);
}
struct Q {
    double val;
    point A;
    int sign;
    Q () {}
    Q (double _val, const point & _A, int _sign) { val = _val; A = _A; sign = _sign; }
    bool operator < (const Q & a) const {
        return val < a.val;
    }
}a[N];
typedef pair<Q, Q> pi;
#define X first
#define Y second
struct ciru {
    double r;
    point o;
    void scan () { scanf ("%lf%lf%lf", &o.x, &o.y, &r); }
    void print () { printf ("%lf %lf %lf\n", o.x, o.y, r); }
    double dis (const ciru & a){
        return (o - a.o).len ();
    }
    bool in (const ciru & a){
        return dcmp (a.r - r - this->dis (a)) >= 0;
    }
    bool out (const ciru & a){
        return dcmp (this->dis (a) - (a.r + r)) >= 0;
    }
    pi G (const ciru & a)
    {
        vec oo = a.o - o;
        double d = oo.len ();
        double cos = (sqr (r) + sqr (d) - sqr (a.r)) / (2 * r * d);
        vec t1 = oo.rot (cos, sqrt (1 - sqr (cos))).pull (r / d);
        vec t2 = oo.rot (cos, -sqrt (1 - sqr (cos))).pull (r / d);
        //printf ("%lf %lf %lf\n", t1.x, t1.y, atan2 (t1.y, t1.x));
        //printf ("%lf %lf %lf\n", t2.x, t2.y, atan2 (t2.y, t2.x));
        //puts ("**************************");
        double t;
        return pi (Q ((t = atan2 (t1.y, t1.x)) < 0 ? t + 2 * PI : t, o + t1, 1), 
                   Q ((t = atan2 (t2.y, t2.x)) < 0 ? t + 2 * PI : t, o + t2, -1));
    }
}cir[N];
inline bool check (int x)
{
    for (int i = 1; i <= n; i ++)
    {
        if (i == x || v[i]) continue;
        if (cir[x].in (cir[i]))
            return true;
    }
    return false;
}
inline double area (Q & a, const Q & b, double r)
{
    double t = b.val - a.val;
    //printf ("%lf\n", sqr (r) * (t - sin (t)) + a.A.cross (b.A));
    return 0.5 * sqr (r) * (t - sin (t)) + 0.5 * a.A.cross (b.A);
}
inline double G (int x)
{ 
    int cnt = 0, now = 0;double S (0);pi tmp;point O;
    a[++ cnt].val = 0;
    O = a[cnt].A = cir[x].o + vec (cir[x].r, 0);
    a[cnt].sign = 0;
    a[++ cnt].val = 2 * PI;
    a[cnt].A = O;
    a[cnt].sign = 0;
    for (int i = 1; i <= n; i ++)
    {
        if (i == x || cir[x].out (cir[i])) continue;
        tmp = cir[x].G (cir[i]);
        a[++ cnt] = tmp.X;
        a[++ cnt] = tmp.Y;
        if (a[cnt - 1].val > a[cnt].val)
            a[++ cnt] = Q (0, O, 1), a[++ cnt] = Q (2 * PI, O, -1);
    }
    sort (a + 1, a + 1 + cnt);
    for (int i = 1; i < cnt; i ++)
    {
        now += a[i].sign;
        if (now == 0)
            S += area (a[i], a[i + 1], cir[x].r);
    }
    return S;
}
int main ()
{
    //freopen ("a.in", "r", stdin);
    //freopen ("a.out", "w", stdout);
    scanf ("%d", &n);
    for (int i = 1; i <= n; i ++)
        cir[i].scan ();
    for (int i = 1; i <= n; i ++)
        if (dcmp (cir[i].r) == 0)
            v[i] = true;
    for (int i = 1; i <= n; i ++)
    {
        if (v[i]) continue;
        v[i] = check (i);
    }
    for (int i = n; i >= 1; i --)
        if (v[i])
        {
            n --;
            for (int j = i + 1; j <= n + 1; j ++)
                cir[j - 1] = cir[j];
        }
    /*printf ("%d\n", n);
    for (int i = 1; i <= n; i ++)
        cir[i].print ();*/
    
    double S (0);
    for (int i = 1; i <= n; i ++)
        S += G (i);
    printf ("%.3lf\n", S);
    return 0;        
}