CodeForces 379F 树的直径 New Year Tree

题意：每次操作新加两个叶子节点，每次操作完以后询问树的直径。

维护树的直径的两个端点U，V，每次计算一下新加进来的叶子节点到U，V两点的距离，如果有更长的就更新。

因为根据树的直径的求法，若出现新的直径，一定是到U或者到V距离最远。

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>

 const int maxn =  + ;
 const int logmaxn = ;

 int n, Q;

 int L[maxn];
 int fa[maxn];
 int anc[maxn][logmaxn];

 void add(int u, int pa)
 {
     fa[u] = pa;
     L[u] = L[pa] + ;
     anc[u][] = pa;
     for(int j = ; ( << j) < n; j++) if(anc[u][j-])
         anc[u][j] = anc[anc[u][j-]][j-];
 }

 int LCA(int p, int q)
 {
     if(L[p] < L[q]) std::swap(p, q);
     int log;
     for(log = ; ( << log) <= L[p]; log++); log--;
     for(int i = log; i >= ; i--)
         if(L[p] - ( << i) >= L[q]) p = anc[p][i];
     if(p == q) return p;
     for(int i = log; i >= ; i--)
         if(anc[p][i] && anc[p][i] != anc[q][i])
             p = anc[p][i], q = anc[q][i];
     return fa[p];
 }

 int distance(int u, int v)
 {
     int l = LCA(u, v);
     return L[u] + L[v] - L[l] * ;
 }

 int main()
 {
     scanf("%d", &Q);
     n = ;
     fa[] = fa[] = fa[] = ;
     L[] = L[] = L[] = ;
     anc[][] = anc[][] = anc[][] = ;

     int U = , V = , diameter = ;
     while(Q--)
     {
         int p; scanf("%d", &p);
         add(++n, p);
         add(++n, p);
         int l1 = distance(n, U), l2 = distance(n, V);
         if(l1 >= l2 && l1 >= diameter)
         {
             V = n;
             diameter = l1;
         }
         else if(l2 >= l1 && l2 >= diameter)
         {
             U = n;
             diameter = l2;
         }
         printf("%d\n", diameter);
     }

     return ;
 }

代码君