poj2420A Star not a Tree?(模拟退火）

链接

求某一点到其它点距离和最小，求这个和，这个点 为费马点。

做法：模拟退火

 1 #include <iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<stdlib.h>
 6 #include<vector>
 7 #include<cmath>
 8 #include<queue>
 9 #include<set>
10 using namespace std;
11 #define N 100000
12 #define LL long long
13 #define INF 0xfffffff
14 const double eps = 1e-8;
15 const double pi = acos(-1.0);
16 const double inf = ~0u>>2;
17 struct point
18 {
19     double x,y;
20     point(double x=0,double y =0):x(x),y(y){}
21 }p[N];
22 int n;
23 int dir[4][2] = {0,1,0,-1,1,0,-1,0};
24 typedef point pointt;
25 pointt operator -(point a,point b)
26 {
27     return point(a.x-b.x,a.y-b.y);
28 }
29 double dis(point a)
30 {
31     return sqrt(a.x*a.x+a.y*a.y);
32 }
33 double cal(point pp)
34 {
35     int i;
36     double s = 0;
37     for(i = 1; i <= n; i++)
38     s+=dis(pp-p[i]);
39     return s;
40 }
41 int main()
42 {
43     int i;
44     while(scanf("%d",&n)!=EOF)
45     {
46         double t = 10000;
47         point pp = point(0,0);
48         for(i =1 ; i <= n;i ++)
49         {
50             scanf("%lf%lf",&p[i].x,&p[i].y);
51             pp.x+=p[i].x;
52             pp.y+=p[i].y;
53         }
54         pp.x /=n;
55         pp.y /=n;
56         double sum = cal(pp),tmp;
57         point tp,ans = pp;
58         while(t>0.02)
59         {
60             int flag = 1;
61             while(flag)
62             {
63                 flag = 0;
64                 for(i =0 ; i< 4 ; i++)
65                 {
66                     tp = point(pp.x+dir[i][0]*t,pp.y+dir[i][1]*t);
67                     tmp = cal(tp);
68                     if(tmp<sum)
69                     {
70                         sum = tmp;
71                         flag = 1;
72                         ans = tp;
73                     }
74                 }
75                 pp = ans;
76             }
77             t/=2;
78         }
79         printf("%.0f\n",sum);
80     }
81     return 0;
82 }

poj2420A Star not a Tree?(模拟退火）