HDU 4305 lightning (Gaussian demetabase Kirchhoff matrix + inverse element)

For some coordinate points, if the distance between the two points is smaller than R and there are no other points between the two points, the two points are connected, which forms a graph. Ask the number of spanning trees in this graph.

Because the data volume is not large, the O (N ^ 3) graph creation is no problem.

After creating the graph, you can use the Kirchhoff matrix to calculate the number of spanning trees. The reason for writing the problem-solving report is that the inverse element is required when Gaussian deyuan solves the Kirchhoff matrix.

(A/B) % mod. If the range of A and B is large, there will be a large error in the result. Here we can convert B * x = 1 (% mod) then X is the reverse element of B (a/B) % mod = (A * X) % mod

The inverse element is the process of solving the linear homogeneous equation B * x reg1 (% MoD ).

For details, see the code:

/* A negative number is not allowed in the whole process of Gaussian element elimination * // # pragma comment (linker, "/Stack: 327680000,327680000 ") # include <iostream> # include <cstdio> # include <cmath> # include <vector> # include <cstring> # include <algorithm> # include <string> # include <set> # include <functional> # include <numeric> # include <sstream> # include <stack> # include <map> # include <queue> # define Cl (ARR, val) memset (ARR, Val, sizeof (ARR) # define rep (I, n) for (I) = 0; (I) <(N); ++ (I) # define for (I, L, H) for (I) = (l); (I) <= (h ); ++ (I) # define Ford (I, H, L) for (I) = (h); (I)> = (l); -- (I )) # define L (x) <1 # define R (x) <1 | 1 # define mid (L, R) (L + r)> 1 # define min (x, y) x <Y? X: y # define max (x, y) x <Y? Y: X # define e (x) (1 <(x) # define iabs (x) <0? -(X): (x) # define out (x) printf ("% i64d \ n", x) # define lowbit (x) & (-x) # define read () freopen ("data. in "," r ", stdin) # define write () freopen (" data. out "," W ", stdout); const double EPS = 1e-8; typedef long ll; const int INF = ~ 0u> 2; using namespace STD; const int n = 330; const int mod = 10007; struct node {Double X; Double Y; double DIS (node CMP) {return (CMP. x-x) * (CMP. x-x) + (CMP. y-y) * (CMP. y-y) ;}} P [N]; int exp_gcd (int A, int B, Int & X, Int & Y) {If (B = 0) {x = 1; y = 0; return a;} int P = exp_gcd (B, A % B, x, y); int TMP = x; X = y; y = TMP-(A/B) * Y; return P;} int det (int A [] [N], int N) {int I, j, k; int ans = 1, x, y; bool flag = true; for (I = 0; I <n; ++ I) {If (! A [I] [I]) {for (j = I + 1; j <n; ++ J) {if (a [J] [I]) {for (k = I; k <n; ++ K) Swap (A [I] [K], a [J] [k]); flag =! Flag; break ;}}if (j = N) Return-1;} ans = (ANS * A [I] [I] % mod + mod) % MOD; exp_gcd (A [I] [I], Mod, x, y); X = (X % mod + mod) % MOD; For (k = I + 1; k <n; ++ k) A [I] [k] = (a [I] [k] * x % mod + mod) % MOD; for (j = I + 1; j <n; ++ J) {if (a [J] [I]) {for (k = I + 1; k <N; ++ K) {A [J] [k] = (A [J] [k]-A [J] [I] * A [I] [k]) % mod + mod) % MOD;} A [J] [I] = 0 ;}} if (FLAG) return ans; Return (-ans % mod + mod) % MOD;} int a [n] [N], D [N] [N], C [N] [N]; int main () {// read (); int t, n, R, I, J, k; CIN> T; while (t --) {CIN> N> r; rep (I, n) CIN> P [I]. x> P [I]. y; CL (A, 0); rep (I, n) {for (j = I + 1; j <n; ++ J) {// printf ("% d % F \ n", I, j, P [I]. DIS (P [J]); If (P [I]. DIS (P [J])> r * r) continue; For (k = 0; k <n; ++ K) {If (k = I | K = J) continue; If (P [K]. y-P [I]. y) * (p [J]. x-P [K]. x) = (P [K]. x-P [I]. x) * (p [J]. y-P [k]. Y) & P [I]. DIS (P [J])> P [I]. DIS (P [k]) & P [I]. DIS (P [J])> P [K]. DIS (P [J]) break;} If (k = n) A [I] [J] = A [J] [I] = 1 ;}} CL (D, 0); bool flag = true; rep (I, n) {d [I] [I] = 0; rep (J, n) {if (a [I] [J]) d [I] [I] ++;} If (d [I] [I] = 0) {flag = false; break;} If (! Flag) {puts ("-1"); continue;} rep (I, n) {rep (J, n) {c [I] [J] = d [I] [J]-A [I] [J]; c [I] [J] = (C [I] [J] % mod + mod) % MOD;} printf ("% d \ n", det (C, n-1);} return 0 ;}