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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
1 /* 3007 */ 2 #include <iostream> 3 #include <sstream> 4 #include <string> 5 #include <map> 6 #include <queue> 7 #include <set> 8 #include <stack> 9 #include <vector> 10 #include <deque> 11 #include <bitset> 12 #include <algorithm> 13 #include <cstdio> 14 #include <cmath> 15 #include <ctime> 16 #include <cstring> 17 #include <climits> 18 #include <cctype> 19 #include <cassert> 20 #include <functional> 21 #include <iterator> 22 #include <iomanip> 23 using namespace std; 24 //#pragma comment(linker,"/STACK:102400000,1024000") 25 26 #define sti set<int> 27 #define stpii set<pair<int, int> > 28 #define mpii map<int,int> 29 #define vi vector<int> 30 #define pii pair<int,int> 31 #define vpii vector<pair<int,int> > 32 #define rep(i, a, n) for (int i=a;i<n;++i) 33 #define per(i, a, n) for (int i=n-1;i>=a;--i) 34 #define clr clear 35 #define pb push_back 36 #define mp make_pair 37 #define fir first 38 #define sec second 39 #define all(x) (x).begin(),(x).end() 40 #define SZ(x) ((int)(x).size()) 41 #define lson l, mid, rt<<1 42 #define rson mid+1, r, rt<<1|1 43 44 typedef struct { 45 double x, y; 46 } Point; 47 48 const double eps = 1e-6; 49 const int maxn = 505; 50 Point P[maxn]; 51 Point o; 52 double r; 53 int n; 54 55 double Length(Point a, Point b) { 56 return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)); 57 } 58 59 double Cross(Point a, Point b, Point c) { 60 return (c.x-a.x)*(b.y-a.y) - (c.y-a.y)*(b.x-a.x); 61 } 62 63 Point Intersect(Point a, Point b, Point c, Point d) { 64 Point ret = a; 65 double t = ((a.x - c.x) * (c.y - d.y) - (a.y - c.y) * (c.x - d.x)) / 66 ((a.x - b.x) * (c.y - d.y) - (a.y - b.y) * (c.x - d.x)); 67 ret.x += (b.x - a.x) * t; 68 ret.y += (b.y - a.y) * t; 69 return ret; 70 } 71 72 Point circumcenter(Point a, Point b, Point c) { 73 Point ua, ub, va, vb; 74 75 ua.x = (a.x + b.x) / 2.0; 76 ua.y = (a.y + b.y) / 2.0; 77 ub.x = ua.x - a.y + b.y; 78 ub.y = ua.y + a.x - b.x; 79 80 va.x = (a.x + c.x) / 2.0; 81 va.y = (a.y + c.y) / 2.0; 82 vb.x = va.x - a.y + c.y; 83 vb.y = va.y + a.x - c.x; 84 return Intersect(ua, ub, va, vb); 85 } 86 87 void min_center() { 88 o = P[0]; 89 r = 0; 90 91 rep(i, 1, n) { 92 if (Length(P[i], o)-r > eps) { 93 o = P[i]; 94 r = 0; 95 rep(j, 0, i) { 96 if (Length(P[j], o)-r > eps) { 97 o.x = (P[i].x + P[j].x) / 2; 98 o.y = (P[i].y + P[j].y) / 2; 99 r = Length(o, P[j]); 100 101 rep(k, 0, j) { 102 if (Length(P[k], o)-r > eps) { 103 o = circumcenter(P[i], P[j], P[k]); 104 r = Length(o, P[k]); 105 } 106 } 107 } 108 } 109 } 110 } 111 } 112 113 void solve() { 114 min_center(); 115 printf("%.2lf %.2lf %.2lf\n", o.x, o.y, r); 116 } 117 118 int main() { 119 ios::sync_with_stdio(false); 120 #ifndef ONLINE_JUDGE 121 freopen("data.in", "r", stdin); 122 freopen("data.out", "w", stdout); 123 #endif 124 125 while (scanf("%d", &n)!=EOF && n) { 126 rep(i, 0, n) 127 scanf("%lf%lf", &P[i].x, &P[i].y); 128 solve(); 129 } 130 131 #ifndef ONLINE_JUDGE 132 printf("time = %d.\n", (int)clock()); 133 #endif 134 135 return 0; 136 }
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原文地址:http://www.cnblogs.com/bombe1013/p/5234419.html