POJ2069：Super Star

我对模拟退火的理解：https://www.cnblogs.com/AKMer/p/9580982.html

我对爬山的理解：https://www.cnblogs.com/AKMer/p/9555215.html

题目传送门：http://poj.org/problem?id=2069

这题和求费马点没啥多大区别……就是把二维改成三维了，然后原本求距离和改成求最远距离了。

如果你不知道费马点是啥的话可以去看看这篇博客：

POJ2420 A Star not a Tree？：https://www.cnblogs.com/AKMer/p/9594350.html

时间复杂度：(O(能A))

空间复杂度：(O(能A))

爬山算法代码如下：

#include <cmath>
#include <ctime>
#include <cstdio>
#include <algorithm>
using namespace std;

#define sqr(a) ((a)*(a))

int n;
double ansx,ansy,ansz,ans;

struct point {
	double x,y,z;
}p[35];

double len() {
	double x=rand()%200000-100000;
	return x/100000;
}

double dis(double x1,double y1,double z1,double x2,double y2,double z2) {
	return sqrt(sqr(x1-x2)+sqr(y1-y2)+sqr(z1-z2));
}

double calc(double x,double y,double z) {
	double tmp=0;
	for(int i=1;i<=n;i++)
		tmp=max(tmp,dis(x,y,z,p[i].x,p[i].y,p[i].z));//改成取max
	if(tmp<ans)ans=tmp;
	return tmp;
}

int main() {
	srand(time(0));
	while(~scanf("%d",&n)) {
		if(!n)break;ans=1e18;
        ansx=0,ansy=0,ansz=0;
		for(int i=1;i<=n;i++) {
			scanf("%lf%lf%lf",&p[i].x,&p[i].y,&p[i].z);
			ansx+=p[i].x,ansy+=p[i].y,ansz+=p[i].z;
		}ansx/=n,ansy/=n,ansz/=n;
		double now_x=ansx,now_y=ansy,now_z=ansz;
		for(double T=1e7;T>=1e-7;T*=0.998) {
			double nxt_x=now_x+len()*T;
			double nxt_y=now_y+len()*T;
			double nxt_z=now_z+len()*T;
			if(calc(nxt_x,nxt_y,nxt_z)<calc(now_x,now_y,now_z))
				now_x=nxt_x,now_y=nxt_y,now_z=nxt_z;
		}printf("%.5lf
",ans);
	}
	return 0;
}

模拟退火代码如下：

#include <cmath>
#include <ctime>
#include <cstdio>
#include <algorithm>
using namespace std;

#define sqr(a) ((a)*(a))

const double T_0=1e-7;
const double del_T=0.998;

int n;
double ansx,ansy,ansz,ans;

struct point {
	double x,y,z;
}p[35];

double len() {
	double x=rand()%200000-100000;
	return x/100000;
}

double dis(double x1,double y1,double z1,double x2,double y2,double z2) {
	return sqrt(sqr(x1-x2)+sqr(y1-y2)+sqr(z1-z2));
}

double calc(double x,double y,double z) {
	double tmp=0;
	for(int i=1;i<=n;i++)
		tmp=max(tmp,dis(x,y,z,p[i].x,p[i].y,p[i].z));
	if(tmp<ans)ans=tmp;
	return tmp;
}

void Anneal() {
	double T=1e7,now_x=ansx,now_y=ansy,now_z=ansz;
	while(T>=T_0) {
		double nxt_x=now_x+len()*T;
		double nxt_y=now_y+len()*T;
		double nxt_z=now_z+len()*T;
		double tmp1=calc(now_x,now_y,now_z);
		double tmp2=calc(nxt_x,nxt_y,nxt_z);
		if(tmp1>tmp2||exp((tmp1-tmp2)/T)*RAND_MAX>rand())
			now_x=nxt_x,now_y=nxt_y,now_z=nxt_z;
		T*=del_T;
	}
}

int main() {
	while(~scanf("%d",&n)) {
		if(!n)break;ans=1e18;
		ansx=0,ansy=0,ansz=0;
		for(int i=1;i<=n;i++) {
			scanf("%lf%lf%lf",&p[i].x,&p[i].y,&p[i].z);
			ansx+=p[i].x,ansy+=p[i].y,ansz+=p[i].z;
		}ansx/=n,ansy/=n,ansz/=n;
		Anneal();printf("%.5lf
",ans);
	}
	return 0;
}

POJ数据好强啊！！！根本A不掉啊！