树上博弈（sg函数）

题目

#include<bits/stdc++.h>
using namespace std;
#define int long long
const int maxn = 2e5 + 5, mod = 1e9 + 7;
vector<int> G[maxn];
int sg[maxn];//sg[i]表示以1为根时，以i为根的子树的sg值
int n, fa[maxn];
int sg2[maxn];//sg2[i]表示以i为根时，整棵树的sg值
void init(){
	for(int i = 1; i <= n; i++){
		sg[i] = sg2[i] = 0;
		G[i].clear();
	}
}
void dfs(int u, int f){
	fa[u] = f;
	for(auto v : G[u]){
		if(v == f) continue;
		dfs(v, u);
		sg[u] ^= (sg[v] + 1);//sg[u] = (sg[v1] + 1) ^ (sg[v2] + 1) ^ ... ^ (sg[vn] + 1)
	}
}
void dfs2(int u, int f){
	for(auto v : G[u]){
		if(v == f) continue;
		sg2[v] = sg[v] ^ ((sg2[u] ^ (sg[v] + 1)) + 1);//换根
		dfs2(v, u);
	}
}
int qpow(int a, int b){
	int res = 1;
	while(b){
		if(b & 1) res = res * a % mod;
		a = a * a % mod;
		b >>= 1;
	}
	return res;
}
void solve(){
	int i, j;
	cin >> n;
	init();
	for(i = 1; i < n; i++){
		int u, v;
		cin >> u >> v;
		G[u].push_back(v);
		G[v].push_back(u);
	}
  dfs(1, 1);
	int cnt = 0, SG;
	sg2[1] = sg[1];
	dfs2(1, 1);
	for(i = 1; i <= n; i++){
		if(sg2[i]) cnt++;
	}
  cout << cnt * qpow(n, mod - 2) % mod << '\n';
}
signed main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	int T;
	cin >> T;
	while(T--){
		solve();
	}
	return 0;
}