ZOJ 1450 HDU 3007 (最小圆覆盖)

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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;
}

ZOJ 1450 HDU 3007 (最小圆覆盖)

标签：

原文地址：http://blog.csdn.net/accelerator_/article/details/44451259