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题目：点击打开链接
题意：最小球覆盖裸题。
分析：三分套三分再套三分（x,y,z三个方向嵌套三分）或者模拟退火。
代码一（三分）：
#include<algorithm>
#include<iostream>
#include<fstream>
#include<complex>
#include<cstdlib>
#include<cstring>
#include<cassert>
#include<iomanip>
#include<string>
#include<cstdio>
#include<bitset>
#include<vector>
#include<cctype>
#include<cmath>
#include<ctime>
#include<stack>
#include<queue>
#include<deque>
#include<list>
#include<set>
#include<map>
using namespace std;
#define pt(a) cout<<a<<endl
#define debug test
#define mst(ss,b) memset((ss),(b),sizeof(ss))
#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 pii pair<int,int>
#define fi first
#define se second
#define ll long long
#define ull unsigned long long
#define pb push_back
#define mp make_pair
#define inf 0x3f3f3f3f
#define eps 1e-10
#define PI acos(-1.0)
const ll mod = 1e9+7;
const int N = 1e2+10;
ll qp(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
int to[4][2]={{-1,0},{1,0},{0,-1},{0,1}};
struct P{
double x,y,z;
}p[N];
double dis(P A,P B){
return sqrt((A.x-B.x)*(A.x-B.x)+(A.y-B.y)*(A.y-B.y)+(A.z-B.z)*(A.z-B.z));
}
int n;
double cal(double x,double y,double z)
{
double sum=0;
for(int i=0;i<n;i++)sum=max(sum,dis(p[i],(P){x,y,z}));
return sum;
}
double cal(double x,double y)
{
double l=-100000,r=100000,ans=1e18;
for(int i=1;i<=50;i++)
{
double mid=(l+r)/2;
double m=(mid+r)/2;
double L=cal(x,y,mid);
double R=cal(x,y,m);
if(L>R)l=mid;
else r=m;
ans=min(ans,L);
ans=min(ans,R);
}
return ans;
}
double cal(double x)
{
double l=-100000,r=100000,ans=1e18;
for(int i=1;i<=50;i++)
{
double mid=(l+r)/2;
double m=(mid+r)/2;
double L=cal(x,mid);
double R=cal(x,m);
if(L>R)l=mid;
else r=m;
ans=min(ans,L);
ans=min(ans,R);
}
return ans;
}
int main()
{
cin>>n;
for(int i=0;i<n;i++)scanf("%lf%lf%lf",&p[i].x,&p[i].y,&p[i].z);
double l=-100000,r=100000,ans=1e18;
for(int i=1;i<=50;i++)
{
double mid=(l+r)/2;
double m=(mid+r)/2;
double L=cal(mid);
double R=cal(m);
if(L>R)l=mid;
else r=m;
ans=min(ans,L);
ans=min(ans,R);
}
printf("%.5f\n",ans);
return 0;
}代码二（退火法）：
///#pragma comment(linker, "/STACK:102400000,102400000")
///#include<unordered_map>
///#include<unordered_set>
#include<algorithm>
#include<iostream>
#include<fstream>
#include<complex>
#include<cstdlib>
#include<cstring>
#include<cassert>
#include<iomanip>
#include<string>
#include<cstdio>
#include<bitset>
#include<vector>
#include<cctype>
#include<cmath>
#include<ctime>
#include<stack>
#include<queue>
#include<deque>
#include<list>
#include<set>
#include<map>
using namespace std;
#define pt(a) cout<<a<<endl
#define debug test
#define mst(ss,b) memset((ss),(b),sizeof(ss))
#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 pii pair<int,int>
#define fi first
#define se second
#define ll long long
#define ull unsigned long long
#define pb push_back
#define mp make_pair
#define inf 0x3f3f3f3f
#define eps 1e-10
#define PI acos(-1.0)
const ll mod = 1e9+7;
const int N = 110;
ll qp(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
int to[4][2]={{-1,0},{1,0},{0,-1},{0,1}};
int n;
struct P{
double x,y,z;
}s,p[N];
double delta,ans;
double dis(P a,P b) {
return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)+(a.z-b.z)*(a.z-b.z));
}
void sa() {
s.x=s.y=s.z=0;
delta=10000,ans=1e20;
///delta开始要设大一点，不然会wa
while(delta>eps) {
int d=1;
for(int i=2;i<=n;i++)
if(dis(s,p[i])>dis(s,p[d])) d=i;
double md=dis(s,p[d]);
ans=min(ans,md);
s.x+=(p[d].x-s.x)/md*delta;
s.y+=(p[d].y-s.y)/md*delta;
s.z+=(p[d].z-s.z)/md*delta;
delta*=0.99;
}
cout<<fixed<<setprecision(7)<<ans<<endl;
}
int main() {
ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
while(cin>>n) {
for(int i=1;i<=n;i++) cin>>p[i].x>>p[i].y>>p[i].z;
sa();
}
return 0;
}