【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-6;
const int maxn = 505;
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[0];
    r = 0;
    
    rep(i, 1, n) {
        if (Length(P[i], o)-r > eps) {
            o = P[i];
            r = 0;
            rep(j, 0, i) {
                if (Length(P[j], o)-r > eps) {
                    o.x = (P[i].x + P[j].x) / 2;
                    o.y = (P[i].y + P[j].y) / 2;
                    r = Length(o, P[j]);
                    
                    rep(k, 0, 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
", 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, 0, n)
            scanf("%lf%lf", &P[i].x, &P[i].y);
        solve();
    }
    
    #ifndef ONLINE_JUDGE
        printf("time = %d.
", (int)clock());
    #endif
    
    return 0;
}