/*
Floyd算法：最短路
关键：
（1）求出矩形第四个点坐标（判断垂直交点，向量加）
已知矩形的A、B、C点的坐标，求D点坐标：
若 |AB|^2 + |AC|^2 = |BC|^2，则：A点即为垂直交点
向量(A->B) + 向量(A->C) = 向量(A->D)
=> D.x = A.x + 向量(A->D).x
D.y = A.y + 向量(A->D).y
（2）建图
*/
1 #include <iostream>
2 #include <cstdlib>
3 #include <cstdio>
4 #include <cstddef>
5 #include <iterator>
6 #include <algorithm>
7 #include <string>
8 #include <locale>
9 #include <cmath>
10 #include <vector>
11 #include <cstring>
12 #include <map>
13 #include <utility>
14 #include <queue>
15 #include <stack>
16 #include <set>
17 #include <functional>
18 using namespace std;
19 typedef pair<double, int> P;
20 const int INF = 0x3f3f3f3f;
21 const int modPrime = 3046721;
22 const double eps = 1e-9;
23 const int MaxN = 105;
24
25 int n, s, A, B;
26 double t;
27 double T[MaxN];
28
29 double dp[MaxN][MaxN];
30
31 struct Point
32 {
33 int x, y;
34 };
35
36 vector<Point> cityPoint[MaxN];
37
38
39 int disSquare(const Point p1, const Point p2)
40 {
41 return (p1.x - p2.x)*(p1.x - p2.x) + (p1.y - p2.y)*(p1.y - p2.y);
42 }
43
44 int verticalPoint(const Point pt[])
45 {
46 int dis01, dis02, dis12;
47 dis01 = disSquare(pt[0], pt[1]);
48 dis02 = disSquare(pt[0], pt[2]);
49 dis12 = disSquare(pt[1], pt[2]);
50
51 if (dis01 + dis02 == dis12) return 0;
52 if (dis01 + dis12 == dis02) return 1;
53 if (dis02 + dis12 == dis01) return 2;
54
55 return -1;
56 }
57
58 void iniCity(int city)
59 {
60 Point pt[4];
61 scanf("%d %d %d %d %d %d %lf", &pt[0].x, &pt[0].y, &pt[1].x, &pt[1].y, &pt[2].x, &pt[2].y, &T[city]);
62
63 int vtcPt = verticalPoint(pt);
64
65 pt[3].x = pt[vtcPt].x + ((pt[(vtcPt + 1) % 3].x - pt[vtcPt].x) + (pt[(vtcPt + 2) % 3].x - pt[vtcPt].x));
66 pt[3].y = pt[vtcPt].y + ((pt[(vtcPt + 1) % 3].y - pt[vtcPt].y) + (pt[(vtcPt + 2) % 3].y - pt[vtcPt].y));
67
68 for (int i = 0; i < 4; ++i)
69 {
70 cityPoint[city].push_back(pt[i]);
71 }
72
73 for (int i = 0; i < 4; ++i)
74 {
75 for (int j = i; j < 4; ++j)
76 {
77 if (i != j)
78 {
79 dp[i + (city << 2)][j + (city << 2)] = T[city] * sqrt(1.0*disSquare(pt[i], pt[j]));
80 dp[j + (city << 2)][i + (city << 2)] = dp[i + (city << 2)][j + (city << 2)];
81 }
82 else
83 {
84 dp[i + (city << 2)][j + (city << 2)] = 0.0;
85 }
86 }
87 }
88 }
89
90 void iniMap()
91 {
92 for (int i = 0; i < s; ++i)
93 {
94 for (int j = i + 1; j < s; ++j)
95 {
96 for (int m = 0; m < 4; ++m)
97 {
98 for (int n = 0; n < 4; ++n)
99 {
100 dp[m + (i << 2)][n + (j << 2)] = t*sqrt(1.0*disSquare(cityPoint[i][m], cityPoint[j][n]));
101 dp[n + (j << 2)][m + (i << 2)] = dp[m + (i << 2)][n + (j << 2)];
102 }
103 }
104 }
105 }
106 }
107
108 void Solve()
109 {
110 int pointSum = (s << 2);
111 for (int k = 0; k < pointSum; ++k)
112 {
113 for (int i = 0; i < pointSum; ++i)
114 {
115 for (int j = 0; j < pointSum; ++j)
116 {
117 dp[i][j] = min(dp[i][j], dp[i][k] + dp[j][k]);
118 }
119 }
120 }
121 }
122
123 void getAns()
124 {
125 double ans = INF * 1.0;
126 for (int i = 0; i < 4; ++i)
127 {
128 for (int j = 0; j < 4; ++j)
129 {
130 ans = min(ans, dp[i + ((A - 1) << 2)][j + ((B - 1) << 2)]);
131 }
132 }
133 printf("%.1lf\n", ans);
134 }
135
136
137 int main()
138 {
139 #ifdef HOME
140 freopen("in", "r", stdin);
141 //freopen("out", "w", stdout);
142 #endif
143
144 scanf("%d", &n);
145 while (n--)
146 {
147 scanf("%d %lf %d %d", &s, &t, &A, &B);
148 for (int i = 0; i < s; ++i)
149 {
150 iniCity(i);
151 }
152 iniMap();
153 Solve();
154 getAns();
155 for (int i = 0; i < s; ++i)
156 {
157 cityPoint[i].clear();
158 }
159 }
160
161 #ifdef HOME
162 cerr << "Time elapsed: " << clock() / CLOCKS_PER_SEC << " ms" << endl;
163 _CrtDumpMemoryLeaks();
164 #endif
165 return 0;
166 }
