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
", cal (l));
50 }
51
52 return 0;
53 }