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 200010int 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;} elsel = mid + 1;}printf("%d %d %d\n", ans[0], ans[1], ans[2]);}return 0;
}