CF917D Stranger Trees(矩阵树定理，多项式求解系数)

题意

给定一个个点完全图，以及一棵树，树上边为关键边，问完全图中恰好含条关键边的生成树有多少。

题解

对于基尔霍夫矩阵树定理，其具体求解的是，我们一般计算题目取便是单纯的生成树计数，这里考虑生成函数相关：我们取关键边的，那么求解后的系数就表示恰好条边生成树个数。

我们可以知道我们通过矩阵树定理的行列式会算出一个的多项式，那么我们可以考虑对不同值进行具体计算，算出个，然后可以通过高斯消元求解出多项式系数。

#include 
#include 
#include 
#include 
#include 
using namespace std;

using LL = long long;
const int MAXN = 1e2 + 5;
const int MOD = 1e9 + 7;
int N;
int mp[MAXN][MAXN];
int A[MAXN][MAXN];
int Guass_Elimination_Det(int n);
int b[MAXN], p[MAXN];
int inv(int a, int p);
int qpow(int x, int n);

int main() {
  //
  scanf("%d", &N);
  for (int i = 1; i < N; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    mp[x][y] = mp[y][x] = 1;
  }
  for (int k = 1; k <= N; k++) {
    for (int i = 1; i <= N; i++) A[i][i] = 0;
    for (int i = 1; i <= N; i++)
      for (int j = 1; j <= N; j++) {
        if (i == j) continue;
        if (mp[i][j])
          A[i][j] = MOD - k, A[i][i] += k;
        else
          A[i][j] = MOD - 1, A[i][i] += 1;
      }
    b[k] = Guass_Elimination_Det(N - 1);
  }
  //
  for (int i = 1; i <= N; i++) {
    A[i][1] = 1;
    for (int j = 2; j <= N; j++) A[i][j] = 1LL * i * A[i][j - 1] % MOD;
  }
  for (int k = 1; k < N; k++)
    for (int i = k + 1; i <= N; i++)
      while (A[i][k]) {
        int tmp = A[k][k] / A[i][k];
        for (int j = k; j <= N; j++) {
          A[k][j] = (A[k][j] - 1LL * tmp * A[i][j] % MOD + MOD) % MOD;
          swap(A[k][j], A[i][j]);
        }
        b[k] = (b[k] - 1LL * tmp * b[i] % MOD + MOD) % MOD;
        swap(b[k], b[i]);
      }
  for (int i = N; i >= 1; i--) {
    p[i] = b[i];
    for (int j = i + 1; j <= N; j++)
      p[i] = (p[i] - 1LL * p[j] * A[i][j] % MOD + MOD) % MOD;
    p[i] = 1LL * inv(A[i][i], MOD) * p[i] % MOD;
  }
  for (int i = 1; i <= N; i++) printf("%d ", p[i]);
  return 0;
}

int Guass_Elimination_Det(int n) {
  int res = 1, mark = 1;
  for (int k = 1; k <= n; k++) {
    if (!A[k][k])
      for (int i = k + 1; i <= n; i++)
        if (A[i][k]) {
          for (int j = k; j <= n; j++) swap(A[k][j], A[i][j]);
          mark *= -1;
          break;
        }
    //
    for (int i = k + 1; i <= n; i++) {
      int tmp = 1LL * A[i][k] * inv(A[k][k], MOD) % MOD;
      for (int j = k + 1; j <= n; j++)
        A[i][j] = (A[i][j] - 1LL * tmp * A[k][j] % MOD + MOD) % MOD;
      A[i][k] = 0;
    }
    if (!A[k][k]) return 0;
    res = 1LL * res * A[k][k] % MOD;
  }
  if (mark == -1) res = MOD - res;
  return res;
}

int qpow(int x, int n) {
  int res = 1;
  while (n) {
    if (n & 1) res = 1LL * res * x % MOD;
    x = 1LL * x * x % MOD;
    n >>= 1;
  }
  return res;
}

int inv(int a, int p) {
  // cout << a << " " << p << endl;
  a %= p;
  if (a == 1) return 1;
  return p - 1LL * p * inv(p, a) / a;
}