Expectation

思路

题目要求每个生成树边权$\&$的期望值，假设当前这颗生成树对二进制数的第$i$位有贡献，则这个位上的构成生成树的边权值一定是$1$，所以我们可以跑$31$位二进制数的，矩阵树，每个位上的贡献度等于，这个位上的生成树数量乘以这个位上的2次幂，最后再跑一边生成树计数，然后即可求得期望。

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[110][110];const int mod = 998244353;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) {return quick_pow(a, mod - 2, mod);
}ll gauss(int n){ll ans = 1;for(int i = 1; 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]); 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;
}const int N = 1e4 + 10;int x[N], y[N], w[N], n, m;int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);int T = read();while(T--) {n = read(), m = read();for(int i = 1; i <= m; i++) {x[i] = read(), y[i] = read(), w[i] = read();}ll ans = 0;for(int i = 0; i <= 30; i++) {memset(A, 0, sizeof A);for(int j = 1; j <= m; j++) {add(x[j], y[j], w[j] >> i & 1);add(y[j], x[j], w[j] >> i & 1);}ans = (ans + (gauss(n - 1) * (1 << i) % mod)) % mod;}memset(A, 0, sizeof A);for(int j = 1; j <= m; j++) {add(x[j], y[j], 1);add(y[j], x[j], 1);}ans = ans * inv(gauss(n - 1)) % mod;printf("%lld\n", (ans % mod + mod) % mod);}return 0;
}