对于求深度和宽度都很好维护。深度dfs时维护就行，宽度统计同一个深度的节点有多少个，然后取max。
对于求距离，我刚开始以为是要走到根节点在回来，然后固输了(dep[u] - 1) * 2 + dep[v] - 1，结果竟然得了80分，数据有点水过头了……
实际上就是求LCA，然而因为只求一次，所以暴力就行啦~~
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<cstdlib>
#include<stack>
#include<queue>
#include<vector>
#include<cctype>
using namespace std;
#define enter puts("")
#define space putchar(' ')
#define Mem(a) memset(a, 0, sizeof(a))
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-;
const int maxn = ;
inline ll read()
{
ll ans = ;
char ch = getchar(), last = ' ';
while(!isdigit(ch)) {last = ch; ch = getchar();}
while(isdigit(ch)) {ans = ans * + ch - ''; ch = getchar();}
if(last == '-') ans = -ans;
return ans;
}
inline void write(ll x)
{
if(x < ) putchar('-'), x = -x;
if(x >= ) write(x / );
putchar(x % + '');
} int n, s, t;
vector<int> v[maxn]; int dep[maxn], fa[maxn];
bool vis[maxn];
void dfs(int now)
{
vis[now] = ;
for(int i = ; i < (int)v[now].size(); ++i)
{
if(!vis[v[now][i]])
{
dep[v[now][i]] = dep[now] + ;
fa[v[now][i]] = now;
dfs(v[now][i]);
}
}
} int cnt[maxn], Max_dep = -, Max_wid = -; int lca(int x, int y)
{
int ret = ;
while(x != y)
{
if(dep[x] >= dep[y]) {x = fa[x]; ret += ;} //一定要有等于，否则同一深度就无限循环了
else if(dep[x] < dep[y]) {y = fa[y]; ret++;}
}
return ret;
} int main()
{
n = read();
for(int i = ; i < n; ++i)
{
int x = read(), y = read();
v[x].push_back(y); v[y].push_back(x);
}
s = read(); t = read();
dep[] = ;
dfs();
for(int i = ; i <= n; ++i)
{
Max_dep = max(Max_dep, dep[i]);
Max_wid = max(Max_wid, ++cnt[dep[i]]);
}
write(Max_dep); enter; write(Max_wid); enter;
write(lca(s, t));
// write(((dep[s] - 1 ) << 1) + dep[t] - 1); enter;
return ;
}