題意

座標軸中有 $n$個點，需要畫一個圓，這個人與 $x$軸相切，且包含全部的 $n$

題解

二分搜尋半徑，觀察可以發現，這個圓的圓心一定是在 $y = R$上面移動的，所以我們可以依據這個半徑確定每個點的圓心在 $y$

程式碼

#include <iostream>
#include <cmath>
using namespace std;
const int maxn = 1e5+5;
const double EPS = 1e-8;
int n;
struct Node {
    double x,y;
}cor[maxn];

bool ok(double x) {
    double l = -1e10, r = 1e10;
    for(int i = 0; i < n; ++i) {
        double h = abs(cor[i].y-x);
        if(h > x) return 0;
        double range = sqrt(cor[i].y*(2*x-cor[i].y));
        l = max(l, cor[i].x-range), r = min(r, cor[i].x+range);
    }
    return (r-l)>EPS;
}
int main() {
    scanf("%d", &n);
    bool fg1 = false, fg2 = false;
    for(int i = 0; i < n; ++i) {
        scanf("%lf%lf", &cor[i].x, &cor[i].y);
        if(cor[i].y < 0)
            fg1 = true, cor[i].y = -cor[i].y;
        else 
            fg2 = true;
    }
    if(fg1 && fg2) {
        puts("-1");
    }
    else if(n == 1) {
        printf("%lf\n", cor[0].y/2);
    }
    else {
        double l = 0, r = 1e18, ans;
        for(int i = 0; i < 1000; ++i) {
            double mid = (l+r)/2;
         //   cout << mid << endl;
            if(ok(mid)) {
                r = mid;
                ans = mid;
            }
            else l = mid;
        }
        printf("%lf\n", ans);
    }
    return 0;
}