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 D[x][x]+=z;D[y][y]+=z;D[x][x]+=z;D[y][y]+=z 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 ) (x,y,z)(x,y,z) (x,y,z)(x,y,z) ,则 C [ x ] [ y ] + = z ; C [ y ] [ x ] + = z ; C [ x ] [ y ] + = z ; C [ y ] [ x ] + = z C[x][y]+=z;C[y][x]+=z;C[x][y]+=z;C[y][x]+=z 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 − C A = D − C A=D−CA=D−C 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。
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 ,则 外向树中 ,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 D[x][x]+=z;D[x][x]+=z D[x][x]+=z;D[x][x]+=z;
邻接矩阵C:若存在边 ( x , y , z ) ( x , y , z ) (x,y,z)(x,y,z) (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 C[x][y]+=z;C[x][y]+=z C[x][y]+=z;C[x][y]+=z;
图G的基尔霍夫矩阵 A = D − C A = D − C A=D−CA=D−C 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。
/*
Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include
#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 | 1
using 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;
}