Weird Flecks, But OK

題意：

給出三維坐標中的 n 個點，求一個圓柱的最小直徑，該圓柱垂直于坐標平面且能覆蓋住所有點

題解：

本人最不擅長計算幾何，比賽時沒做出來。。。
其實就是將n個點投影到三個坐標平面，在三個坐標面上分別做期望時間復雜度為O(n)的最下圓覆蓋算法，最后取最小值
將三維轉為二維問題

代碼：

#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #include<vector> #include<set> #include<map> #define ll long long #define inf 0x3f3f3f3f #define Inf 0x3f3f3f3f3f3f3f3f #define int ll using namespace std; int read() {int res = 0,flag = 1;char ch = getchar();while(ch<'0' || ch>'9'){if(ch == '-') flag = -1;ch = getchar();}while(ch>='0' && ch<='9'){res = (res<<3)+(res<<1)+(ch^48);//res*10+ch-'0';ch = getchar();}return res*flag; } const int maxn = 2e5+5; const int maxm = 3e4+5; const int mod = 998244353; const double pi = acos(-1); const double eps = 1e-8; int n; struct Point{double x,y; }a[maxn],b[maxn],c[maxn]; double get_dis(Point a,Point b) {return hypot(a.x-b.x,a.y-b.y); } void updat(Point p1,Point p2,Point p3,Point &o,double &r) //用確定的三點更新外接圓圓心和半徑 {double xx1 = 2*(p2.x-p1.x),yy1 = 2*(p2.y-p1.y);double c1 = p2.x*p2.x+p2.y*p2.y-p1.x*p1.x-p1.y*p1.y;double xx2 = 2*(p3.x-p1.x),yy2 = 2*(p3.y-p1.y);double c2 = p3.x*p3.x+p3.y*p3.y-p1.x*p1.x-p1.y*p1.y;o.x = (yy1*c2-yy2*c1)/(yy1*xx2-yy2*xx1);o.y = (xx2*c1-xx1*c2)/(yy1*xx2-yy2*xx1);r=get_dis(o,p1); } double get_circle(Point a[]) //求能覆蓋a中所有點的最小外接圓直徑 {random_shuffle(a+1,a+n+1);Point o = a[1];double r = 0;for(int i = 2;i <= n;i++)if(get_dis(o,a[i]) > r+eps){o = a[i],r = 0;for(int j = 1;j < i;j++)if(get_dis(o,a[j]) > r+eps){o.x = (a[i].x+a[j].x)/2;o.y = (a[i].y+a[j].y)/2;r = get_dis(a[i],a[j])/2;for(int k = 1;k < j;k++)if(get_dis(o,a[k]) > r+eps) updat(a[i],a[j],a[k],o,r);}}return 2*r; } signed main() {n = read();for(int i = 1;i <= n;i++){double x,y,z;scanf("%lf%lf%lf",&x,&y,&z);a[i] = {x,y};b[i] = {y,z};c[i] = {x,z};}double ans;ans = min({get_circle(a),get_circle(b),get_circle(c)});printf("%.10f\n",ans);return 0; }

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