Steiner tree

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const int MAXN = 105;
 5 const double INF = 1e12;
 6 pair<int, int> p[MAXN];
 7 double dp[MAXN][1 << 11], dis[MAXN][MAXN];
 8 int N2(int x) {
 9         return x * x;
10 }
11 double get_dis(pair<int, int> a, pair<int, int> b) {
12         return sqrt(N2(a.first - b.first) + N2(a.second - b.second));
13 }
14 int main() {
15         int n, k;
16         scanf("%d %d", &n, &k);
17         for (int i = 0; i < n; i++) {
18                 scanf("%d %d", &p[i].first, &p[i].second);
19         }
20         for (int i = 0; i < n; i++) {
21                 for (int j = 0; j < n; j++) {
22                         if (i != j) {
23                                 dis[i][j] = get_dis(p[i], p[j]);
24                         }
25                 }
26         }
27         for (int i = k; i < n; i++) {
28                 for (int j = 0; j < (1 << k); j++) {
29                         dp[i][j] = 0;
30                         for (int s = 0; s < k; s++) {
31                                 dp[i][j] += (j >> s & 1) * dis[i][s];
32                         }
33                 }
34         }
35         for (int S = 0; S < (1 << k); S++) {
36                 for (int x = k; x < n; x++) {
37                         for (int i = k; i < n; i++) {
38                                 if (i != x) {
39                                         for (int sub = S; ; sub = (sub - 1)&S) {
40                                                 dp[x][S] = min(dp[x][S], dis[i][x] + dp[x][sub] + dp[i][S ^ sub]);
41                                                 if (!sub) {
42                                                         break;
43                                                 }
44                                         }
45                                 }
46                         }
47                 }
48         }
49         double ans = INF;
50         for (int i = k; i < n; i++) {
51                 ans = min(ans, dp[i][(1 << k) - 1]);
52         }
53         printf("%.5f\n", ans);
54         return 0;
55 }