ZOJ 1450 Minimal Circle 最小圆覆盖

来源：互联网发布：全站简繁体转换js代码编辑：程序博客网时间：2024/05/16 11:52

经典的计算几何题目，最小圆覆盖。

最小的圆肯定落在三个点上，因此暴力枚举圆上的三个点即可。

点增量算法，复杂度O(n^3)

加入随机化，平均复杂度可以降到O(n^2)

三点的外接圆圆心的函数：

POINT circumcenter(POINT &a, POINT &b, POINT &c) {    POINT ret;    double a1=b.x-a.x, b1=b.y-a.y, c1=(a1*a1+b1*b1)/2;    double a2=c.x-a.x, b2=c.y-a.y, c2=(a2*a2+b2*b2)/2;    double d = a1*b2 - a2*b1;    ret.x = a.x + (c1*b2-c2*b1)/d;    ret.y = a.y + (a1*c2-a2*c1)/d;    return ret;}

这道题目在HDU上面也有，但是两道题目代码不能互相AC，应该是数据的原因，各种精度问题，还是看重算法吧，暂且把数据无视了。

#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <cstdlib>using namespace std;#define N 110struct POINT {    double x, y;} p[N];int n;inline double dist(const POINT &a, const POINT &b) {    return sqrt((a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y));}POINT circumcenter(POINT &a, POINT &b, POINT &c) {    POINT ret;    double a1=b.x-a.x, b1=b.y-a.y, c1=(a1*a1+b1*b1)/2;    double a2=c.x-a.x, b2=c.y-a.y, c2=(a2*a2+b2*b2)/2;    double d = a1*b2 - a2*b1;    ret.x = a.x + (c1*b2-c2*b1)/d;    ret.y = a.y + (a1*c2-a2*c1)/d;    return ret;}void solve() {    random_shuffle(p, p+n);  //随机化序列，std里面的随机函数    POINT c;    double r = 0;    for (int i=1; i<n; i++) {        if (dist(p[i], c) <= r) continue;        c = p[i];        r = 0;        for (int j=0; j<i; j++) {            if (dist(p[j], c) <= r) continue;            c.x = (p[i].x+p[j].x)/2;            c.y = (p[i].y+p[j].y)/2;            r = dist(p[j], c);            for (int k=0; k<j; k++) {                if (dist(p[k], c) <= r) continue;                c = circumcenter(p[i], p[j], p[k]);                r = dist(p[i], c);            }        }    }    printf("%.2lf %.2lf %.2lf\n", c.x, c.y, r);}int main() {    while (scanf(" %d", &n) == 1 && n) {        for (int i=0; i<n; i++)            scanf(" %lf %lf", &p[i].x, &p[i].y);        solve();    }    return 0;}