1 /*
2 题意：中文题
3
4 题解：赤裸裸的最小生成树
5 */
6 #include <cstdio>
7 #include <cmath>
8 #include <algorithm>
9
10 const int MAXN = 109;
11 const int MAXE = 10009;
12
13 struct edge
14 {
15 int u, v;
16 double w;
17 bool vis;
18 }E[MAXE];
19
20 int set[MAXN];
21 int N, EN, sum;
22
23 void addedge(int u, int v, double w)
24 {
25 E[ EN ].u = u;
26 E[ EN ].v = v;
27 E[ EN ].w = w;
28 E[ EN ].vis = false;
29 EN++;
30 return ;
31 }
32
33 void init()
34 {
35 for (int i = 1; i <= N; i++) // 树的标号从1开始
36 set[i] = i;
37 return ;
38 }
39
40 int find(int x)
41 {
42 while (set[x] != x)
43 x = set[x];
44 return set[x];
45 }
46
47 bool cmp(edge a, edge b)
48 {
49 return a.w < b.w;
50 }
51
52 double kruskal()
53 {
54 int x, y;
55 double ret;
56 init();
57 sum = 0;
58 ret = 0;
59 std::sort(E,E+EN,cmp);
60 for (int i = 0; i < EN && sum < N - 1; i++)
61 {
62 x = find( E[i].u );
63 y = find( E[i].v );
64 if (x == y)
65 continue;
66 set[x] = set[y];
67 E[i].vis = true;
68 sum++;
69 ret += E[i].w;
70 }
71 return ret;
72 }
73
74 struct point
75 {
76 int x,y;
77 }P[MAXN];
78
79 int getdisSq(point a, point b)
80 {
81 return (a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y);
82 }
83
84 int main(void)
85 {
86 int t;
87 scanf("%d",&t);
88 while (t--)
89 {
90 scanf("%d",&N);
91 for(int i=1; i<=N; i++)
92 scanf("%d%d",&P[i].x,&P[i].y);
93 EN = 0;
94 for(int i=1; i<N; i++)
95 {
96 for(int j=i+1; j<=N; j++)
97 {
98 int tmp = getdisSq(P[i],P[j]);
99 if (100 <= tmp && tmp <= 1000000)
100 addedge(i,j,sqrt(1.0*tmp));
101 }
102 }
103 double ans = kruskal();
104 if (sum == N-1)
105 printf("%.1lf\n",100*ans);
106 else
107 printf("oh!\n");
108 }
109 return 0;
110 }
