ZOJ 2928 Mathematical contest in modeling

题意是求一个到所给点距离之和小的点。
模仿超哥，用退火的方法。（详细算法解释请看http://www.cnblogs.com/heaad/archive/2010/12/20/1911614.html）

#include<iostream>
#include<cstring>
#include<string>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<set>
#include<map>
#include<queue>
using namespace std;

#define inf 0x3f3f3f3f
#define eps 1e-8
#define maxn 105
#define mod 1000007
#define FOR(i,s,t) for(int i = s; i < t; i++ )
#define LL long long int
#define ULL unsigned long long
#define pii pair<int,int>

//FILE *fin = fopen( "data.in","r" );

struct point {
 double x,y,z;
 point () {};
 point ( double xx,double yy,double zz ): x(xx),y(yy),z(zz) {};

 point operator - ( const point &b ) const {
  return point( x-b.x,y-b.y, z-b.z);
 }

 double len (){
  return sqrt( x*x+ y*y + z*z );
 }
}p[maxn];
int n, dx[30],dy[30],dz[30];

double cal ( point a ){          //计算点a到所有点的距离之和
 double ans = 0;
 FOR(i,0,n)
  ans += (a-p[i]).len();
 return ans;
}

int main (){
 int cnt = 0;
 FOR(i,-1,2)
  FOR(j,-1,2)
  FOR( k,-1,2)
  dx[cnt] = i, dy[cnt] = j, dz[cnt] = k, ++cnt;        //当前点的移动方向

 while( ~scanf("%d", &n) ){
  FOR(i,0,n)
   scanf("%lf%lf%lf",&p[i].x, &p[i].y, &p[i].z);
  point cur(0,0,0), nxt;
  double dis = cal(cur);         //dis要满足最优解在其范围之内
  double ans = dis;
  while( dis > eps ){        
   FOR(i,0,cnt){             //满足精度退出
    nxt = point(cur.x + dx[i]*dis, cur.y + dy[i]*dis, cur.z + dz[i]*dis );
    double t = cal(nxt);
    if( t < ans )          //找到局部最优解，更新
     ans = t, cur = nxt;
   }
   dis *= 0.99;
  }
  printf("%.3lf %.3lf %.3lf\n",cur.x,cur.y,cur.z);
 }

 return 0;
}