poj 2420 求n个点的费马点 随机算法解决

还是跟那题poj 1379 run away一样的做法

http://www.cnblogs.com/wuyiqi/archive/2011/12/08/2280885.html

改一下最优的方法（到n个点的距离之和）就好了

//求n边形的费马点
//即找到一个点使得这个点到n个点的距离之和最小
#include<stdio.h>
#include<math.h>
#include<stdlib.h>
#include<time.h>
const double inf = 1e10;
const double pi = acos(-1.0);
const int Rp = 4;//初始时随机选择一些点，不用太多
const int shift = 60;//但是方向一定要多
struct point {
    double x,y;
    void goto_rand_dir(double key)
    {
        double d=2*pi*(double)rand()/RAND_MAX;
        x+=key*sin(d);
        y+=key*cos(d);
    }
    void Get_Rand_Point(int a,int b)
    {
        x=rand()%a+1;
        y=rand()%b+1;
    }
}p[1010],randp[Rp];
double Dis(point a,point b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
double dis[Rp];
int main()
{
    int i,j,x,y,k,m;
    srand(time(NULL));
    while(scanf("%d",&m)!=EOF)
    {
        for(i=0;i<m;i++)
        {
            scanf("%lf%lf",&p[i].x,&p[i].y);
        }
        x=10000;y=10000;
        double tmp;
        for(i=0;i<Rp;i++)
        {
            dis[i]=inf;
            randp[i].Get_Rand_Point(x,y);
            tmp=0;
            for(j=0;j<m;j++)
            {
                tmp+=Dis(randp[i],p[j]);    
            }
            if(tmp<dis[i])
                dis[i]=tmp;
        }
        double key=sqrt(1.0*(x*x+y*y));//初始的步长，要保证初始点从该点出发肯定能包括整个区域
        while(key>=0.01)
        {
            for(i=0;i<Rp;i++)
            {
                for(j=0;j<shift;j++)
                {
                    point cc=randp[i];
                    cc.goto_rand_dir(key);
                    if(cc.x<0||cc.y<0||cc.x>x||cc.y>y) continue;
                    tmp=0;
                    for(k=0;k<m;k++)
                    {
                        tmp+=Dis(cc,p[k]);
                    }
                    if(tmp<dis[i]) //如果从i点出发随机移动的点比原来的点更优，则接受该移动
                    {
                        dis[i]=tmp;
                        randp[i]=cc;
                    }
                }
            }
            key=key*0.6;//可灵活调整
        }
        for(i=k=0;i<Rp;i++)
            if(dis[i]<dis[k])
                k=i;
            printf("%.0lf\n",dis[k]);
    }
    return 0;
}

网上的另一种做法，一个点朝四个方向按一定方向搜索，然后再逐步缩小步长

参考自http://hi.baidu.com/perfectcai_/blog/item/65349a48840bc4e382025cc2.html

代码写的简单多了，但这种做法有时候会错。。因为只有四个方向，还是上面的做法保险

也贴贴代码吧

#include <iostream>
#include <cmath>
#define N 128
#define EPS 0.1
using namespace std;
int n;
double x[N], y[N];
double f(double a, double b)
{
    int i;
    double ans = 0;
    for (i=0; i<n; i++)
        ans += sqrt((a-x[i])*(a-x[i]) + (b-y[i])*(b-y[i]));

    return ans;
}
int main()
{
    double ans, step, temp, a, b, ta, tb;
    int i;
    bool flag;
    while (scanf("%d", &n) == 1)
    {
        for (i=0; i<n; i++)
            scanf("%lf%lf", &x[i], &y[i]);
        a = x[0];
        b = y[0];
        ans = f(a, b);
        step = 100;
        while (step > EPS)
        {
            flag = 1;
            while (flag)
            {
                flag = 0;
                temp = f(a+step, b);
                if (temp < ans)
                {
                    flag = 1;
                    ans = temp;
                    ta = a + step;
                    tb = b;
                }
                temp = f(a-step, b);
                if (temp < ans)
                {
                    flag = 1;
                    ans = temp;
                    ta = a - step;
                    tb = b;
                }
                temp = f(a, b+step);
                if (temp < ans)
                {
                    flag = 1;
                    ans = temp;
                    ta = a;
                    tb = b + step;
                }
                temp = f(a, b-step);
                if (temp < ans)
                {
                    flag = 1;
                    ans = temp;
                    ta = a;
                    tb = b - step;
                }
                a = ta;
                b = tb;
            }
            step /= 2;
        }
        printf("%.0lf\n", ans);
    }
    return 0;
}