三分 POJ 2420 A Star not a Tree?

 

题目传送门

 1 /*
 2     题意：求费马点
 3     三分：对x轴和y轴求极值，使到每个点的距离和最小
 4 */
 5 #include <cstdio>
 6 #include <algorithm>
 7 #include <cstring>
 8 #include <cmath>
 9 
10 const int MAXN = 1e2 + 10;
11 const int INF = 0x3f3f3f3f;
12 double x[MAXN], y[MAXN];
13 int n;
14 
15 double sum(double x1, double y1)    {
16     double res = 0;
17     for (int i=1; i<=n; ++i)    {
18         res += sqrt ((x1 - x[i]) * (x1 - x[i]) + (y1 - y[i]) * (y1 - y[i]));
19     }
20     return res;
21 }
22 
23 double cal(double x1)   {
24     int l = 0.0, r = 10000.0;
25     for (int i=1; i<=100; ++i)  {
26         double lmid = (l + r) / 2;
27         double rmid = (lmid + r) / 2;
28         if (sum (x1, lmid) < sum (x1, rmid))    r = rmid;
29         else    l = lmid;
30     }
31     return sum (x1, l);
32 }
33 
34 int main(void)  {       //POJ 2420 A Star not a Tree?
35     //freopen ("POJ_2420.in", "r", stdin);
36 
37     while (scanf ("%d", &n) == 1)   {
38         for (int i=1; i<=n; ++i)    {
39             scanf ("%lf%lf", &x[i], &y[i]);
40         }
41 
42         double l = 0.0, r = 10000.0;
43         for (int i=1; i<=100; ++i)  {
44             double lmid = (l + r) / 2;
45             double rmid = (lmid + r) / 2;
46             if (cal (lmid) < cal (rmid))    r = rmid;
47             else    l = lmid;
48         }
49         printf ("%.0f\n", cal (l));
50     }
51 
52     return 0;
53 }

 

posted @ 2015-07-25 18:50  Running_Time  阅读(164)  评论(0)  编辑  收藏  举报