Bzoj1336 Bzoj1337 最小圆覆盖

来源：互联网发布：java中文本框取值编辑：程序博客网时间：2024/05/16 13:44

题意简述：
给出 N (2≤n≤105) 个点，让你画一个最小的包含所有点的圆。
题解：
http://blog.csdn.net/acm_cxlove/article/details/7887305（不能写得更好）

吐槽：老年选手算两直线交点叉积两边元素放反调死了才调出来 ……

#include<bits/stdc++.h>const int N = 1e5 + 10;const double eps = 1e-12;const double Pi = acos(-1);struct P {    double x, y;    P() {}    P(double _x, double _y):        x(_x), y(_y) {}};double sqr(double x) {return x * x;}P operator + (const P &a, const P &b) {return P(a.x + b.x, a.y + b.y);}P operator - (const P &a, const P &b) {return P(a.x - b.x, a.y - b.y);}P operator * (const P &a, double base) {return P(a.x * base, a.y * base);};double operator * (const P &a, const P &b) {return a.x * b.y - b.x * a.y;}double dot(const P &a, const P &b) {return a.x * b.x + a.y * b.y;}double dis(const P &a, const P &b) {return sqrt(sqr(a.x - b.x) + sqr(a.y - b.y));};P mid(const P &a, const P &b) {return P((a.x + b.x) * 0.5, (a.y + b.y) * 0.5);};P rotate(const P &p, double alpha) {return P(cos(alpha) * p.x - sin(alpha) * p.y, sin(alpha) * p.x + cos(alpha) * p.y);}struct L {    P p, v;    L() {}    L(P _p, P _v):        p(_p), v(_v) {}};P cross(const L &a, const L &b) {return a.p + a.v * (b.v * (a.p - b.p) / (a.v * b.v));}struct C {    P o;    double r;    C() {}    C(P _o, double _r):        o(_o), r(_r) {}};C get(const P &a, const P &b) {return C(mid(a, b), dis(a, b) * 0.5);}C get(const P &a, const P &b, const P &c) {    if (abs((b - a) * (c - a)) < eps) {        double d1 = dis(a, b), d2 = dis(b, c), d3 = dis(c, a);        if (d1 > d2 && d1 > d3) return get(a, b);        if (d2 > d3) return get(b, c); else return get(c, a);    };    P t1 = rotate(c - a, Pi / 2), t2 = rotate(c - b, Pi / 2);    P o = cross(L(mid(a, c), rotate(c - a, Pi / 2)), L(mid(b, c), rotate(c - b, Pi / 2)));    return C(o, dis(o, c));}bool inclr(const P &p, const C &o) {return dis(o.o, p) < o.r + eps;}int n;P p[N];int main() {    scanf("%d", &n);    for (int i = 1; i <= n; i++)        scanf("%lf%lf", &p[i].x, &p[i].y);    std::random_shuffle(p + 1, p + n + 1);    C ans = C(p[1], 0);    for (int i = 2; i <= n; i++){        if (inclr(p[i], ans)) continue;        ans = C(p[i], 0);        for (int j = 1; j < i; j++) {            if (inclr(p[j], ans)) continue;            ans = C(mid(p[i], p[j]), dis(p[i], p[j]) * 0.5);            for (int k = 1; k < j; k++) {                if (inclr(p[k], ans)) continue;                ans = get(p[i], p[j], p[k]);            }        }    }    printf("%.6f\n", ans.r);    printf("%.6f %.6f\n", ans.o.x, ans.o.y);    return 0;}

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