【HDOJ】3007 Buried memory

1. 题目描述
有n个点，求能覆盖这n个点的半径最小的圆的圆心及半径。

2. 基本思路
算法模板http://soft.cs.tsinghua.edu.cn/blog/?q=node/1066定义Di表示相对于P[1]和P[i]组成的最小覆盖圆，如果P[2..i-1]都在这个圆内，那么当前的圆心和半径即为最优解。
如果P[j]不在这个圆内，那么P[j]一定在新的最小覆盖圆的边界上即P[1]、P[j]、P[i]组成的圆。
因为三点可以确定一个圆，因此只需要不断的找到不满足的P[j]，进而更新最优解即可。
其实就是三层循环，不断更新最优解。然而，这个算法的期望复杂度是O（n）。这个比较难以理解。

3. 代码

 /* 3007 */
 #include <iostream>
 #include <sstream>
 #include <string>
 #include <map>
 #include <queue>
 #include <set>
 #include <stack>
 #include <vector>
 #include <deque>
 #include <bitset>
 #include <algorithm>
 #include <cstdio>
 #include <cmath>
 #include <ctime>
 #include <cstring>
 #include <climits>
 #include <cctype>
 #include <cassert>
 #include <functional>
 #include <iterator>
 #include <iomanip>
 using namespace std;
 //#pragma comment(linker,"/STACK:102400000,1024000")

 #define sti                set<int>
 #define stpii            set<pair<int, int> >
 #define mpii            map<int,int>
 #define vi                vector<int>
 #define pii                pair<int,int>
 #define vpii            vector<pair<int,int> >
 #define rep(i, a, n)     for (int i=a;i<n;++i)
 #define per(i, a, n)     for (int i=n-1;i>=a;--i)
 #define clr                clear
 #define pb                 push_back
 #define mp                 make_pair
 #define fir                first
 #define sec                second
 #define all(x)             (x).begin(),(x).end()
 #define SZ(x)             ((int)(x).size())
 #define lson            l, mid, rt<<1
 #define rson            mid+1, r, rt<<1|1

 typedef struct {
     double x, y;
 } Point;

 const double eps = 1e-;
 const int maxn = ;
 Point P[maxn];
 Point o;
 double r;
 int n;

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

 double Cross(Point a, Point b, Point c) {
     return (c.x-a.x)*(b.y-a.y) - (c.y-a.y)*(b.x-a.x);
 }

 Point Intersect(Point a, Point b, Point c, Point d) {
     Point ret = a;
     double t = ((a.x - c.x) * (c.y - d.y) - (a.y - c.y) * (c.x - d.x)) /
                ((a.x - b.x) * (c.y - d.y) - (a.y - b.y) * (c.x - d.x));
     ret.x += (b.x - a.x) * t;
     ret.y += (b.y - a.y) * t;
     return ret;
 }

 Point circumcenter(Point a, Point b, Point c) {
     Point ua, ub, va, vb;

     ua.x = (a.x + b.x) / 2.0;
     ua.y = (a.y + b.y) / 2.0;
     ub.x = ua.x - a.y + b.y;
     ub.y = ua.y + a.x - b.x;

     va.x = (a.x + c.x) / 2.0;
     va.y = (a.y + c.y) / 2.0;
     vb.x = va.x - a.y + c.y;
     vb.y = va.y + a.x - c.x;
     return Intersect(ua, ub, va, vb);
 }

 void min_center() {
     o = P[];
     r = ;

     rep(i, , n) {
         if (Length(P[i], o)-r > eps) {
             o = P[i];
             r = ;
             rep(j, , i) {
                 if (Length(P[j], o)-r > eps) {
                     o.x = (P[i].x + P[j].x) / ;
                     o.y = (P[i].y + P[j].y) / ;
                     r = Length(o, P[j]);

                     rep(k, , j) {
                         if (Length(P[k], o)-r > eps) {
                             o = circumcenter(P[i], P[j], P[k]);
                             r = Length(o, P[k]);
                         }
                     }
                 }
             }
         }
     }
 }

 void solve() {
     min_center();
     printf("%.2lf %.2lf %.2lf\n", o.x, o.y, r);
 }

 int main() {
     ios::sync_with_stdio(false);
     #ifndef ONLINE_JUDGE
         freopen("data.in", "r", stdin);
         freopen("data.out", "w", stdout);
     #endif

     while (scanf("%d", &n)!=EOF && n) {
         rep(i, , n)
             scanf("%lf%lf", &P[i].x, &P[i].y);
         solve();
     }

     #ifndef ONLINE_JUDGE
         printf("time = %d.\n", (int)clock());
     #endif

     return ;
 }