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2 years ago
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
#define int long long
#define endl "\n"
vector<int> g[N]; // 邻接表,存图
int sz[N]; // sz[i]:以i为根的子树中节点个数
int son[N]; // son[i]:去掉节点i后剩下的连通分量中最大子树节点个数
int r1, r2, n;
void dfs(int u, int fa) {
sz[u] = 1; // u为根的子树中最起码有一个节点u
son[u] = 0; // 把节点u去掉后剩下的连通分量中最大子树节点个数现在还不知道预求最大先设最小
for (int i = 0; i < g[u].size(); i++) { // 枚举u的每一条出边
int v = g[u][i];
if (v == fa) continue;
dfs(v, u);
sz[u] += sz[v];
son[u] = max(son[u], sz[v]);
}
son[u] = max(son[u], n - sz[u]);
if ((son[u] << 1) <= n) r2 = r1, r1 = u;
}
signed main() {
int T;
cin >> T;
while (T--) {
cin >> n;
// 多组测试数据,清空
memset(sz, 0, sizeof sz);
memset(son, 0, sizeof son);
r1 = r2 = 0;
for (int i = 0; i <= n + 10; i++) g[i].clear();
for (int i = 1; i < n; i++) { // n-1条边
int x, y;
cin >> x >> y;
g[x].push_back(y);
g[y].push_back(x);
}
dfs(1, 0); // 以1号点为入口它的父节点是0
if (!r2) {
int r3 = g[r1][0];
cout << r1 << " " << r3 << endl;
cout << r1 << " " << r3 << endl;
} else {
int r3 = r1;
for (int i = 0; i < g[r2].size(); i++) {
r3 = g[r2][i];
if (r3 != r1) break;
}
cout << r3 << " " << r2 << endl;
cout << r3 << " " << r1 << endl;
}
}
return 0;
}