poj 2069 Super Star 模拟退火
题目大意:
给定三位空间上的n(\(n \leq 30\))个点,求最小的球覆盖掉所有的点.
题解:
貌似我们可以用类似于二维平面中的随机增量法瞎搞一下
但是我不会怎么搞
所以我们模拟退火就好了啊QAQ
#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
inline void read(int &x){
x=0;char ch;bool flag = false;
while(ch=getchar(),ch<'!');if(ch == '-') ch=getchar(),flag = true;
while(x=10*x+ch-'0',ch=getchar(),ch>'!');if(flag) x=-x;
}
inline int cat_max(const int &a,const int &b){return a>b ? a:b;}
inline int cat_min(const int &a,const int &b){return a<b ? a:b;}
const int maxn = 45;
const double eps = 1e-15;
const double det = 0.99;
struct Point{
double x,y,z;
Point(const double &a=0,const double &b=0,const double &c=0){
x=a;y=b;z=c;
}
};
inline double dis(const Point &a,const Point &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 ans = 1e100;
Point p[maxn];
inline double f(const Point &x){
double ret = 0;
for(int i=1;i<=n;++i) ret = max(ret,dis(x,p[i]));
if(ret < ans) ans = ret;
return ret;
}
inline double ran(){
return (rand() % 1000 + 1)/1000.0;
}
Point nw;
int main(){
srand(2333);
while(1){
read(n);if(n == 0) break;
nw.x = nw.y = nw.z = 0;
ans = 1e100;
for(int i=1;i<=n;++i){
scanf("%lf%lf%lf",&p[i].x,&p[i].y,&p[i].z);
}
double T = 100.0,x;
double dist;int pos;
while(T > eps){
dist = 0.0;
for(int i=1;i<=n;++i){
if(dist < dis(nw,p[i])){
pos = i;dist = dis(nw,p[i]);
}
}
Point nx(
nw.x+(p[pos].x-nw.x)/dist*T,
nw.y+(p[pos].y-nw.y)/dist*T,
nw.z+(p[pos].z-nw.z)/dist*T
);
x = f(nw) - f(nx);
if(x > 0 || exp(x/T) > ran()) nw = nx;
T *= det;
}
printf("%.5lf\n",ans);
}
getchar();getchar();
return 0;
}
人就像命运下的蝼蚁,谁也无法操控自己的人生.