n个点求出最小圆覆盖所有点
退火算法不会，不过这题可以用三分套三分写
x轴y轴z轴各三分
1 #include <cstdio>
2 #include <cstring>
3 #include <queue>
4 #include <cmath>
5 #include <algorithm>
6 #include <set>
7 #include <iostream>
8 #include <map>
9 #include <stack>
10 #include <string>
11 #include <vector>
12 #define pi acos(-1.0)
13 #define eps 1e-8
14 #define fi first
15 #define se second
16 #define rtl rt<<1
17 #define rtr rt<<1|1
18 #define bug printf("******\n")
19 #define mem(a,b) memset(a,b,sizeof(a))
20 #define name2str(x) #x
21 #define fuck(x) cout<<#x" = "<<x<<endl
22 #define f(a) a*a
23 #define sf(n) scanf("%d", &n)
24 #define sff(a,b) scanf("%d %d", &a, &b)
25 #define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
26 #define sffff(a,b,c,d) scanf("%d %d %d %d", &a, &b, &c, &d)
27 #define pf printf
28 #define FRE(i,a,b) for(i = a; i <= b; i++)
29 #define FREE(i,a,b) for(i = a; i >= b; i--)
30 #define FRL(i,a,b) for(i = a; i < b; i++)+
31 #define FRLL(i,a,b) for(i = a; i > b; i--)
32 #define FIN freopen("data.txt","r",stdin)
33 #define gcd(a,b) __gcd(a,b)
34 #define lowbit(x) x&-x
35 using namespace std;
36 typedef long long LL;
37 typedef unsigned long long ULL;
38 const int mod = 1e9 + 7;
39 const int maxn = 2e5 + 10;
40 const int INF = 0x3f3f3f3f;
41 const LL INFLL = 0x3f3f3f3f3f3f3f3fLL;
42 int n;
43 struct node {
44 double p[3];
45 } qu[maxn];
46 double cal ( node a ) {
47 double cnt = 0;
48 for ( int i = 0 ; i < n ; i++ )
49 cnt = max ( cnt, sqrt ( ( a.p[0] - qu[i].p[0] ) * ( a.p[0] - qu[i].p[0] ) + ( a.p[1] - qu[i].p[1] ) * ( a.p[1] - qu[i].p[1] ) + ( a.p[2] - qu[i].p[2] ) * ( a.p[2] - qu[i].p[2] ) ) );
50 return cnt;
51 }
52 node check ( int cnt, node now ) {
53 if ( cnt >= 3 ) return now;
54 node ans1, ans2, ans, temp1, temp2;
55 double l = -100000, r = 100000, ll, rr;
56 ans = temp1 = temp2 = now;
57 while ( eps < r - l ) {
58 ll = ( 2 * l + r ) / 3, rr = ( 2 * r + l ) / 3;
59 temp1.p[cnt] = ll, temp2.p[cnt] = rr;
60 ans1 = check ( cnt + 1, temp1 );
61 ans2 = check ( cnt + 1, temp2 );
62 if ( cal ( ans1 ) > cal ( ans2 ) ) ans = ans1, l = ll;
63 else r = rr, ans = ans2;
64 }
65 return ans;
66 }
67 int main() {
68 while ( ~sf ( n ) ) {
69 for ( int i = 0 ; i < n ; i++ ) scanf ( "%lf%lf%lf", &qu[i].p[0], &qu[i].p[1], &qu[i].p[2] );
70 node ans;
71 printf ( "%lf\n", cal ( check ( 0, ans ) ) );
72 }
73 return 0;
74 }
来源：oschina
链接：https://my.oschina.net/u/4265175/blog/3746958