luogu P6178 【模板】Matrix-Tree 定理

1.无向图

假设现在给定一个图 G。

度数矩阵D:若存在边$ (x,y,z)(x,y,z)$ ,则 $D[x][x]+=z;D[y][y]+=z;D[x][x]+=z;D[y][y]+=z$;

邻接矩阵C:若存在边 $(x,y,z)(x,y,z)$ ,则$C[x][y]+=z;C[y][x]+=z;C[x][y]+=z;C[y][x]+=z$;

图G的基尔霍夫矩阵 $A=D-CA=D-C$。

删去任意一行和任意一列，求剩下的矩阵行列式即可。

2.有向图

假设现在给定一个图G.

度数矩阵D:若存在边$ (x,y,z)(x,y,z) $,则外向树中$D[y][y]+=z;D[y][y]+=z$; 内向树中 $D[x][x]+=z;D[x][x]+=z$;

邻接矩阵C:若存在边 $(x,y,z)(x,y,z)$ ,则 内向树和外向树中均为$C[x][y]+=z;C[x][y]+=z$;

图G的基尔霍夫矩阵 $A=D-CA=D-C$。

删去指定的根所在的行和列，求剩下的矩阵行列式即可。

代码

/*Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>#define mp make_pair
#define pb push_back
#define endl '\n'
#define mid (l + r >> 1)
#define lson rt << 1, l, mid
#define rson rt << 1 | 1, mid + 1, r
#define ls rt << 1
#define rs rt << 1 | 1using namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;inline ll read() {ll f = 1, x = 0;char c = getchar();while(c < '0' || c > '9') {if(c == '-')    f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x;
}ll A[310][310];const int mod = 1e9 + 7;ll quick_pow(ll a, ll n, ll mod) {ll ans = 1;while(n) {if(n & 1) ans = ans * a % mod;a = a * a % mod;n >>= 1;}return ans;
}ll inv(ll a, ll n) {return quick_pow(a, mod - 2, mod);
}ll gauss(int n){ll ans = 1;for(int i = 2; i <= n; i++){for(int j = i; j <= n; j++) {if(A[j][i]){for(int k = i; k <= n; k++) swap(A[i][k], A[j][k]);if(i != j) ans = -ans;break;}}if(!A[i][i]) return 0;for(ll j = i + 1, iv = inv(A[i][i], mod); j <= n; j++) {ll t = A[j][i] * iv % mod;for(int k = i; k <= n; k++)A[j][k] = (A[j][k] - t * A[i][k] % mod + mod) % mod;}ans = (ans * A[i][i] % mod + mod) % mod;}return ans;
}void add(int x, int y, int w) {(A[x][y] -= w) %= mod;(A[y][y] += w) %= mod;
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);int n = read(), m = read(), t = read();for(int i = 1; i <= m; i++) {int x = read(), y = read(), w = read();if(!t) {add(x, y, w);add(y, x, w);}else {add(x, y, w);}}printf("%lld\n", gauss(n));return 0;
}