题链

退火法

#include <iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#define eps 1e-7
#define MS 50
using namespace std;

int n,m;
struct point3D {
	double x,y,z;
} data[MS];

double dis(point3D a,point3D 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));
}

double solve() {
	double step=100,ans=1e30,mt;
	point3D z;
	z.x=z.y=z.z=0;
	int s=0;
	while(step>eps) {
		for(int i=0; i<n; i++)
			if(dis(z,data[s])<dis(z,data[i])) s=i;
		mt=dis(z,data[s]);
		ans=min(ans,mt);
		z.x+=(data[s].x-z.x)/mt*step;
		z.y+=(data[s].y-z.y)/mt*step;
		z.z+=(data[s].z-z.z)/mt*step;
		step*=0.98;
	}
	return ans;
}

int main() {
	double ans;
	while(~scanf("%d",&n),n) {
		for(int i=0; i<n; i++)
			scanf("%lf%lf%lf",&data[i].x,&data[i].y,&data[i].z);
		ans=solve();
		printf("%.5f\n",ans);
	}
	return 0;
}

标签：覆盖,int,double,最小,step,POJ.2069,include,point3D,define
来源： https://www.cnblogs.com/Tecode/p/14853585.html