n個點求出最小圓覆蓋所有點

退火演算法不會，不過這題可以用三分套三分寫

x軸y軸z軸各三分 

 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 }