hdu6097

201708 （ACM）计算几何----圆反演

• 总结：做计算几何题目时，能用简单方法就不要用复杂的，应该尽最大可能避开精度问题。

#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pb push_back
#define sz(a) (int)a.size()
#define mp make_pair
#define rep(i, a, b) for(int i=(a); i<(b); i++)
#define de(a) cout<<#a<<"="<<a<<endl;
typedef long long ll;
typedef double db;
typedef pair<int, int> pii;
typedef vector<int> vi;
//--------
const db eps = 1e-8;
int sign(db x) {
    return (x > eps) - (x < -eps);
}
db r;
struct P {
    db x,y;
    P() {}
    P(db x,db y) {
        this -> x = x;
        this -> y = y;
    }
    P operator + (const P &c) const {
        return P(x + c.x, y + c.y);
    }
    P operator - (const P &c) const {
        return P(x - c.x, y - c.y);
    }
    P operator * (const db &c) const {
        return P(x * c, y * c);
    }
    P operator / (const db &c) const {
        return P(x / c, y / c);
    }
};
P read() {
    db x,y;scanf("%lf%lf",&x,&y);
    return P(x,y);
}
db dis(P a) { 
    return sqrt(a.x*a.x+a.y*a.y);
}
db dis(P a, P b) {
    return dis(a - b);
}
P fanyan(P a) {
    db d = dis(a);
    db t = r * r / d / d;
    return a * t;
}
int main() {
    int T;scanf("%d", &T);
    while(T--) {
        scanf("%lf", &r);
        P a = read(), b = read();
        if(sign(dis(a)) <= 0) {
            printf("%.7f\n", r*2.0);
            continue;
        }
        P _a = fanyan(a), _b = fanyan(b);
        db d = dis((_a + _b) / 2);
        db ans = 0;
        if(sign(d - r) > 0) {
            P p1 = (a + b) / 2;
            db t = r / dis(p1);
            p1 = p1 * t;
            ans = dis(p1, a) * 2;
        } else {
            ans = dis(_a - _b);
            ans = ans * dis(a) / r;
        }
        printf("%.7f\n",ans);
    }
    return 0;
}