题意：

树上找两个点  使得其它点到这两点随意一点的距离的最大值最小

思路：

最大值最小  想到二分  在二分的基础上判定这个最大值是否可能

怎样判定这个问题就是怎样选那两个点的问题  非常明显  我们要处理的是直径（不然没意义  最长的就是直径）  那么既然已经有了一个要判定的值x  最好还是就选择距离直径两端点距离为x的点就好

直径上的点最多n个  算上二分的复杂度  O(nlogn)能够解决

代码：

#include<cstdio>

#include<iostream>

#include<cstring>

#include<string>

#include<algorithm>

#include<map>

#include<set>

#include<vector>

#include<queue>

#include<cstdlib>

#include<ctime>

#include<cmath>

using namespace std;

typedef long long LL;

#define N 200010

int t, n;

struct edge {

	int v, next;

} ed[N * 2];

int head[N], tot;

int dis[N], pre[N];

int route[N], m;

void add(int u, int v) {

	ed[tot].v = v;

	ed[tot].next = head[u];

	head[u] = tot++;

}

int qu[N];

int bfs(int u) {

	int l = 0, r = 1;

	int pos = u, val = 0;

	dis[u] = 0;

	qu[0] = u;

	while (l < r) {

		u = qu[l++];

		for (int i = head[u]; ~i; i = ed[i].next) {

			int v = ed[i].v;

			if (dis[v] == -1) {

				pre[v] = u;

				dis[v] = dis[u] + 1;

				qu[r++] = v;

				if (dis[v] > val) {

					val = dis[v];

					pos = v;

				}

			}

		}

	}

	return pos;

}

int flag[N];

bool yes(int ans) {

	memset(flag, 0, sizeof(flag));

	memset(dis, -1, sizeof(dis));

	bfs(route[ans]);

	for (int i = 1; i <= n; i++) {

		if (dis[i] <= ans)

			flag[i] = 1;

	}

	memset(dis, -1, sizeof(dis));

	bfs(route[m - 1 - ans]);

	for (int i = 1; i <= n; i++) {

		if (dis[i] <= ans)

			flag[i] = 1;

	}

	for (int i = 1; i <= n; i++) {

		if (!flag[i])

			return false;

	}

	return true;

}

int main() {

	int i, u, v;

	scanf("%d", &t);

	while (t--) {

		tot = 0;

		memset(head, -1, sizeof(head));

		scanf("%d", &n);

		for (i = 1; i < n; i++) {

			scanf("%d%d", &u, &v);

			add(u, v);

			add(v, u);

		}

		memset(dis, -1, sizeof(dis));

		u = bfs(1);

		memset(dis, -1, sizeof(dis));

		memset(pre, -1, sizeof(pre));

		v = bfs(u);

		m = 0;

		while (v != -1) {

			route[m++] = v;

			v = pre[v];

		}

		int l = 0, r = m - 1, mid, ans[3];

		while (l <= r) {

			mid = (l + r) / 2;

			if (yes(mid)) {

				ans[0] = mid;

				ans[1] = route[mid];

				ans[2] = route[m - 1 - mid];

				if (ans[1] == ans[2]) {

					for (i = 0; i < m; i++) {

						if (route[i] != ans[1]) {

							ans[2] = route[i];

							break;

						}

					}

				}

				r = mid - 1;

			} else

				l = mid + 1;

		}

		printf("%d %d %d\n", ans[0], ans[1], ans[2]);

	}

	return 0;

}