Matrix-tree theorem of bzoj2467 spanning tree

http://www.lydsy.com/JudgeOnline/problem.php?id=2467

Do not know why, see this problem decisively think of maxtrix-tree theorem, and then make Kirchhoff matrix.

But first I need to understand the sample, test 3 kinds of situations to understand the sample.

The example is that two Pentagon laps share an edge, all the points of the center polygon are in degrees 4, and the surrounding +1,-1 have edges and accumulate 1 (for 2 of cases do not write a special sentence).

Then YY processing each Pentagon the remaining 3 points, and then I listed (the code may be a bit long)

And then just abandon the last line of a column, the determinant of the value is good.

(PS: I forgot to take the mold, forgot to take the mold, forgot to take the mold ~ ~ ~ ~ and Kirchhoff matrix has negative numbers, I forgot to make this ~~~~~ the last step of the test 100 is a negative number, and then remembered that did not turn it into a positive (determinant transformation may become negative ) and then +p after mod p is over, 233~~~)

#include <cstdio> #include <iostream> #include <cstring> #include <cstdlib> #include <
algorithm> #include <queue> #include <cmath> using namespace std;
int T;
int n;
int f[1020][1020];
int cnt;
	void work () {for (int. i=1;i<=cnt;i++) {for (int j=1;j<=cnt;j++) f[i][j]= (f[i][j]+2007)%2007;
	} int det=1;
				for (int i=1;i<=cnt;i++) {to (int j=i+1;j<=cnt;j++) {while (f[j][i]) {int t=-f[i][i]/f[j][i];
				for (int k=i;k<=cnt;k++) f[i][k]= (f[i][k]+t*f[j][k])%2007;
				for (int k=i;k<=cnt;k++) swap (f[i][k],f[j][k]);
			Det=-det;
			}} if (f[i][i]==0) {printf ("0\n");
		return;
		} Det*=f[i][i];
	det%=2007;
	} det+=2007;
	det%=2007;
printf ("%d\n", Det);
	} int main () {scanf ("%d", &t);
		while (t--) {memset (f,0,sizeof (f));
		scanf ("%d", &n);
	    Cnt=n;
		    for (int i=1;i<=n;i++) {f[i][i]+=4;
		    f[i][(i+1<=n) i+1:1]+=-1;
	    f[i][(i-1) >=1?i-1:n]+=-1; } for (int i=1;i<=n;i++) {cnt++;
	    	F[i][cnt]=f[cnt][i]+=-1;
	    	f[cnt][cnt]+=2;
	    	F[cnt][cnt+1]+=-1;
	    	
	    	F[cnt+1][cnt]+=-1;
	    	cnt++;
	    	f[cnt][cnt]+=2;
	    	F[cnt][cnt+1]+=-1;
	    	
	    	F[cnt+1][cnt]+=-1;
	    	cnt++;
	    	f[cnt][cnt]+=2;
	    	f[cnt][(i+1) <=n?i+1:1]+=-1;
	    f[(i+1) <=n?i+1:1][cnt]+=-1;
	    } cnt--;
	Work ();
} return 0; }