ZOJ 1450 HDU 3007 (最小圆覆盖)

首先这个圆边上必有至少两点，打乱数组，然后利用枚举，不断重新定义圆，找出最小的圆

代码：

#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>using namespace std;const int N = 100005;const double eps = 1e-8;int n;struct Point {    double x, y;    Point() {}    Point(double x, double y) {        this->x = x;        this->y = y;    }    void read() {        scanf("%lf%lf", &x, &y);    }} p[N];struct Line {    Point a, b;    Line() {}    Line(Point a, Point b) {        this->a = a;        this->b = b;    }};typedef Point Vector;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 * (Vector A, double p) {    return Vector(A.x * p, A.y * p);}Vector operator / (Vector A, double p) {    return Vector(A.x / p, A.y / p);}double Cross(Vector A, Vector B) {return A.x * B.y - A.y * B.x;}int dcmp(double x) {    return x < -eps ? -1 : x > eps;}double dis(Point a, Point b) {    double dx = a.x - b.x;    double dy = a.y - b.y;    return sqrt(dx * dx + dy * dy);}Point GetLineIntersection(Point P, Vector v, Point Q, Vector w) {    Vector u = P - Q;    double t = Cross(w, u) / Cross(v, w);    return P + v * t;}Point CircumcircleOfTriangle(Point a, Point b, Point c) {    Line u,v;    u.a.x = (a.x + b.x) / 2;    u.a.y = (a.y + b.y) / 2;    u.b.x = u.a.x + (u.a.y - a.y);    u.b.y = u.a.y - (u.a.x - a.x);    v.a.x = (a.x + c.x) / 2;    v.a.y = (a.y + c.y) / 2;    v.b.x = v.a.x + (v.a.y - a.y);    v.b.y = v.a.y - (v.a.x - a.x);    return GetLineIntersection(u.a, u.b - u.a, v.a, v.b - v.a);}void Mincir() {    random_shuffle(p, p + n);    Point cir = p[0]; double r = 0;    for (int i = 1; i < n; i++) {        if (dcmp(dis(cir, p[i]) - r) <= 0) continue;        cir = p[i]; r = 0;        for (int j = 0; j < i; j++) {            if(dcmp(dis(cir, p[j]) - r) <= 0) continue;            cir.x = (p[i].x + p[j].x) / 2;            cir.y = (p[i].y + p[j].y) / 2;            r = dis(cir, p[j]);            for (int k = 0; k < j; k++) {                if (dcmp(dis(cir, p[k]) - r) <= 0) continue;                cir = CircumcircleOfTriangle(p[i], p[j], p[k]);                r = dis(cir, p[k]);            }        }    }    printf("%.2f %.2f %.2f\n", cir.x, cir.y, r);}int main() {    while (~scanf("%d", &n) && n) {        for (int i = 0; i < n; i++)            p[i].read();        Mincir();    }    return 0;}