题意：根据两点的距离不大于R，而且中间没有点建立一个图。求生成树计数。
题解：生成图计数+判断点在线段上
oi-wiki索引：Kirchhoff 矩阵树定理
则无向图G的生成树个数 = Kirchhoff 矩阵L 的 n-1 阶主子式的行列式的绝对值。
#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<algorithm>
#include<queue>
#include<stack>
#include<cmath>
#include<vector>
#include<fstream>
#include<set>
#include<map>
#include<sstream>
#include<iomanip>
#define ll long long
using namespace std;
const int MOD = 10007;
int INV[MOD];
const double eps = 1e-8;
const double inf = 1e20;
const double pi = acos(-1.0);
const int maxp = 1010;
//Compares a double to zero
int sgn(double x) {
if (fabs(x) < eps) return 0;
if (x < 0) return -1;
else return 1;
}
//square of a double
inline double sqr(double x) {
return x * x; }
//POINT
struct Point {
double x, y;
Point() {
}
Point(double _x, double _y) {
x = _x;
y = _y;
}
Point operator -(const Point& b)const {
return Point(x - b.x, y - b.y);
}
//叉积
double operator ^(const Point& b)const {
return x * b.y - y * b.x;
}
//点积
double operator *(const Point& b)const {
return x * b.x + y * b.y;
}
Point operator *(const double& k)const {
return Point(x * k, y * k);
}
};
//LINE
struct Line {
Point s, e;
Line() {
}
Line(Point _s, Point _e) {
s = _s;
e = _e;
}
// 点在线段上的判断
bool pointonseg(Point p) {
return sgn((p - s) ^ (e - s)) == 0 && sgn((p - s) * (p - e)) <= 0;
}
};
//求ax = 1( mod m) 的x值，就是逆元(0<a<m)
long long inv(long long a, long long m) {
if (a == 1)return 1;
return inv(m % a, m) * (m - m / a) % m;
}
struct Matrix {
int mat[330][330];
void init() {
memset(mat, 0, sizeof(mat));
}
int det(int n) {
//求行列式的值模上MOD，需要使用逆元
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++)
mat[i][j] = (mat[i][j] % MOD + MOD) % MOD;
int res = 1;
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++)
if (mat[j][i] != 0) {
for (int k = i; k < n; k++)
swap(mat[i][k], mat[j][k]);
if (i != j)
res = (-res + MOD) % MOD;
break;
}
if (mat[i][i] == 0) {
res = -1;//不存在(也就是行列式值为0)
break;
}
for (int j = i + 1; j < n; j++) {
//int mut = (mat[j][i]*INV[mat[i][i]])%MOD;//打表逆元
int mut = (mat[j][i] * inv(mat[i][i], MOD)) % MOD;
for (int k = i; k < n; k++)
mat[j][k] = (mat[j][k] - (mat[i][k] * mut) % MOD + MOD) % MOD;
}
res = (res * mat[i][i]) % MOD;
}
return res;
}
};
int ccw(int x1, int y1, int x2, int y2, int x3, int y3) {
int v = (x2 - x1) * (y3 - y1) - (x3 - x1) * (y2 - y1);
if (v == 0) return 0;
return 1;
}
int t, n, r, x[333], y[333], g[333][333];
int ck(int i, int j) {
if ((x[i] - x[j]) * (x[i] - x[j]) + (y[i] - y[j]) * (y[i] - y[j]) > r * r) return 0;
for (int k = 0; k < n; k++) {
if (k == i || k == j) continue;
Line l = Line(Point(x[i], y[i]), Point(x[j], y[j]));
if (l.pointonseg(Point(x[k], y[k]))) return 0;
}
return 1;
}
int main() {
for (int i = 1; i < MOD; i++) INV[i] = inv(i, MOD);
scanf("%d", &t);
while (t--) {
memset(g, 0, sizeof(g));
scanf("%d%d", &n, &r);
for (int i = 0; i < n; i++) scanf("%d%d", &x[i], &y[i]);
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
if (!ck(i, j)) continue;
g[i][j] = g[j][i] = 1;
}
}
Matrix ma;
ma.init();
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++)
if (i != j && g[i][j]) {
ma.mat[i][j] = -1;
ma.mat[i][i]++;
}
printf("%d\n", ma.det(n - 1));
}
return 0;
}