【模板】最小圆覆盖

P1742 最小圆覆盖

题目描述

给出N个点,让你画一个最小的包含所有点的圆。

输入格式

先给出点的个数N,2<=N<=100000,再给出坐标Xi,Yi.(-10000.0<=xi,yi<=10000.0)

输出格式

输出圆的半径,及圆心的坐标,保留10位小数

输入输出样例

输入 #1
6
8.0 9.0
4.0 7.5
1.0 2.0
5.1 8.7
9.0 2.0
4.5 1.0
 
输出 #1
5.0000000000
5.0000000000 5.0000000000

说明/提示

5.00 5.00 5.0

随机数处理复杂度为O(n)

第一个点定为圆心;

第二个点和第一个点的连线的中点定位圆心,距离为直径;

第三个点要和前两个点进行三点定圆;

#pragma GCC optimize(2)
#pragma GCC optimize(3, "Ofast", "inline")

#include<bits/stdc++.h>//O(n)

#define ll long long
#define met(a, x) memset(a,x,sizeof(a))
#define inf 0x3f3f3f3f
#define ull unsigned long long
using namespace std;
const double eps = 1e-12;
struct node {
    double x, y;
} s[500005];
node o;//圆心坐标
double ri;//半径

double dis(node a, node b) {
    return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}

void getr(node p1, node p2, node p3) {//三个点求三角形圆心坐标和半径
    double a, b, c, d, e, f;
    a = p2.y - p1.y;
    b = p3.y - p1.y;
    c = p2.x - p1.x;
    d = p3.x - p1.x;
    f = p3.x * p3.x + p3.y * p3.y - p1.x * p1.x - p1.y * p1.y;
    e = p2.x * p2.x + p2.y * p2.y - p1.x * p1.x - p1.y * p1.y;
    o.x = (a * f - b * e) / (2 * a * d - 2 * b * c);
    o.y = (d * e - c * f) / (2 * a * d - 2 * b * c);
    ri = dis(o, p1);
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++) {
        cin >> s[i].x >> s[i].y;
    }
    random_shuffle(s + 1, s + n + 1);
    o = s[1];
    ri = 0;
    for (int i = 2; i <= n; i++) {
        if (dis(s[i], o) > ri + eps) {
            o = s[i];
            ri = 0;//第一个点为圆心
            for (int j = 1; j < i; j++) {
                if (dis(o, s[j]) > ri + eps) {
                    o.x = (s[i].x + s[j].x) / 2;
                    o.y = (s[i].y + s[j].y) / 2;
                    ri = dis(o, s[j]);//第一个点和第二个点中点为圆心,距离为直径
                    for (int k = 1; k < j; k++) {
                        if (dis(o, s[k]) > ri + eps) {
                            getr(s[i], s[j], s[k]);//三点定圆
                        }
                    }
                }
            }
        }
    }
    cout << fixed << setprecision(10) << ri << endl;
    cout << fixed << setprecision(10) << o.x << ' ' << fixed << setprecision(10) << o.y << endl;
    return 0;
}

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转载自www.cnblogs.com/nublity/p/11620792.html
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