[JLOI2009]二叉树问题

嘟嘟嘟

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 ;
 }