【HDU4305】Lightning（生成树计数）

There are N robots standing on the ground (Don‘t know why. Don‘t know how). 

Suddenly the sky turns into gray, and lightning storm comes! Unfortunately, one of the robots is stuck by the lightning!

So it becomes overladen. Once a robot becomes overladen, it will spread lightning to the near one.


The spreading happens when: 
  Robot A is overladen but robot B not.
  The Distance between robot A and robot B is no longer than R.
  No other robots stand in a line between them.
In this condition, robot B becomes overladen. 

We assume that no two spreading happens at a same time and no two robots stand at a same position. 


The problem is: How many kind of lightning shape if all robots is overladen? The answer can be very large so we output the answer modulo 10007. If some of the robots cannot be overladen, just output -1. 

There are several cases.
The first line is an integer T (T <= 20), indicate the test cases.
For each case, the first line contains integer N ( 1 <= N <= 300 ) and R ( 0 <= R <= 20000 ), indicate there stand N robots; following N lines, each contains two integers ( x, y ) ( -10000 <= x, y <= 10000 ), indicate the position of the robot. 

One line for each case contains the answer.

#include 
#include 
#include 
#include 
#define sqr(x) ((x)*(x))
using namespace std;
const int M=10007;
const int N=301;  
int inv[M],mat[N][N];
void init(){//求逆元
	inv[1]=1;
	for(int i=2;ii;k--)//该行每列减去第i列的值*d
			c[j][k]=(c[j][k]-c[i][k]*inv[c[i][i]]%M*c[j][i]%M+M)%M;
	}
	return (ans*(w&1?-1:1)+M)%M;
}
struct point{
	int x,y;
}p[N];
int same(point a,point b,point c){//判断是否共线
	return (a.x-c.x)*(b.y-c.y)==(b.x-c.x)*(a.y-c.y)
	&&min(a.x,c.x)<=b.x&&max(a.x,c.x)>=b.x
	&&min(a.y,c.y)<=b.y&&max(a.y,c.y)>=b.y;
}
int main(){
	init();
	int t,n,r;
	scanf("%d",&t);
	while(t--){
		memset(mat,0,sizeof mat);
		scanf("%d%d",&n,&r);
		for(int i=0;i