Spanning Tree lighting final count decided to use this template!

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4 #include <cmath>
 5 #define sqr(x) ((x)*(x))
 6 using namespace std;
 7 const int M=10007;
 8 const int N=301;
 9 int inv[M],mat[N][N];
10 void init(){//求逆元
11     inv[1]=1;
12     for(int i=2; I <M; I ++ )
 13 is          INV [I] = (the MM / I) * INV [M% I]% M;
 14  }
 15  int DET ( int C [] [N], int n-) { // find matrix c n-th ordered sequence of master-type absolute value 
16      int I, J, K, W = 0 , ANS = . 1 ;
 . 17      for (I = . 1 ; I <= n; I ++ )
 18 is      for (J = . 1 ; J <= n-; J ++) C [I] [J] = (C [I] [J]% M + M)% M;
 . 19      for (I = . 1 ; I <= n-; I ++ ) {
 20 is          for (J = I; J <= n-; J ++) // find the i-th row from the first column is not 0 row i of 
21             IF (C [I] [J]) BREAK ;
 22 is          IF (I =! J)
 23 is              the swap (C [I], C [J]);
 24          ANS = ANS * C [I] [I]% M;
 25          for (j = i + . 1 ; j <= n-; j ++) // j th row, i th column becomes 0 
26 is          for (n-K + = . 1 ; K> i; K--) // the row of each column is subtracted column D value i * 
27              C [J] [K] = (C [J] [K] -C [i] [K] * INV [C [i] [i]] *% M C [J] [ I]% M + M)% M;
 28      }
 29      return ANS;
 30  }
 31 is  struct Point {
 32      int X, Y;
 33 is }p[N];
34 int same(point a,point b,point c){   //判断是否共线
35     return (a.x-c.x)*(b.y-c.y)==(b.x-c.x)*(a.y-c.y)
36     &&min(a.x,c.x)<=b.x&&max(a.x,c.x)>=b.x
37     &&min(a.y,c.y)<=b.y&&max(a.y,c.y)>=b.y;
38 }
39 int main(){
40     init();
41     int t,n,r;
42     scanf("%d",&t);
43     while(t--){
44         memset(mat,0,sizeof mat);
45         scanf("%d%d",&n,&r);
46         for(int i=1;i<=n;i++)
47             scanf("%d%d",&p[i].x,&p[i].y);
48         for(int i=1;i<=n;i++)
49         for(int j=i+1;j<=n;j++)
50             if(sqrt(sqr(p[i].x-p[j].x)+sqr(p[i].y-p[j].y))<=r){//距离不大于r
51                 int ok=1;
52                 for(int k=1;k<=n;k++)
53                     if(k!=i&&k!=j&&same(p[i],p[k],p[j]))
54                         ok=0;
55                 if(ok){//构造Kirchhoff矩阵
56                     mat[i][j]=mat[j][i]=-1;
57                     mat[i][i]++;mat[j][j]++;
58                 }
59             }
60         int ans=det(mat,n-1);
61         printf("%d\n", years years: - 1 );
62      }
 63 }