Uva10766 (organising the organisation) generates the tree count-matrix-tree Theorem

/** Question address: * http://uva.onlinejudge.org/index.php? Option = com_onlinejudge & Itemid = 8 & page = show_problem & problem = 1707; * In order to improve the company's work efficiency, the Board of Directors decided to reclassify all employees. * except for one General Manager, all other employees have only one direct supervisor; * Due to the direct interpersonal relationship between employees, A and B may be unwilling to lead the other party directly. * n employees in the company 1 ~ N number, and the board of directors has decided to assign K employees to the general manager. * Jimmy's task is to have a different number of employee classification schemes ;** Algorithm Thought: * if there is no conflict between A and B, there will be an edge between them. * The relationship graph between the employees finally obtained is a spanning tree of the source image; * Although the root of the generative tree is specified, the number of undirected graph spanning trees is irrelevant to the root. * You only need to use the matrix-tree theorem to calculate the number of original graph spanning trees; ** matrix-tree theorem: the number of all different spanning trees of * g is equal to the absolute value of any n-1 primary-child type of the Kirchhoff matrix C [g; * The n-1 primary sub-formula is the new matrix obtained after removing the r row and R column of C [g] For r (1 ≤ r ≤ n, using CR [g]; **/# include <iostream> # include <cstdio> # include <cstring> # include <cmath> # include <algorithm> using namespace STD; const int n = 55; typedef long ll; int d [N] [N]; ll C [N] [N]; // Kirchhoff matrix ll det (ll a [] [N], int N) // generate tree count: matrix-tree theorem {ll ret = 1; for (INT I = 1; I <n; I ++) {for (Int J = I + 1; j <N; j ++) while (A [J] [I]) {ll T = A [I] [I]/A [J] [I]; for (int K = I; k <n; k ++) A [I] [k] = (a [I] [k]-A [J] [k] * t); For (int K = I; k <N; k ++) Swap (A [I] [K], a [J] [k]); ret =-ret ;} if (A [I] [I] = 0) return 0; ret = RET * A [I] [I];} If (Ret <0) ret =-ret; return ret;} int main () {// freopen ("C: \ Users \ admin Istrator \ Desktop \ kd.txt "," r ", stdin); int n, m, K; while (~ Scanf ("% d", & N, & M, & K) {memset (C, 0, sizeof (c); memset (D, 0, sizeof (d); int U, V; while (M --) {scanf ("% d", & U, & V); U --; V --; d [u] [v] = d [v] [u] = 1 ;}for (INT I = 0; I <n; I ++) {int u = 0; for (Int J = 0; j <n; j ++) {if (I! = J &&! D [I] [J]) {u ++; C [I] [J] =-1 ;}} C [I] [I] = u ;} ll res = det (C, N); printf ("% LLD \ n", Res);} return 0 ;}