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 = 1000000 + 1000;
const int logmaxn = 20;

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] + 1;
    anc[u][0] = pa;
    for(int j = 1; (1 << j) < n; j++) if(anc[u][j-1])
        anc[u][j] = anc[anc[u][j-1]][j-1];
}

int LCA(int p, int q)
{
    if(L[p] < L[q]) std::swap(p, q);
    int log;
    for(log = 1; (1 << log) <= L[p]; log++); log--;
    for(int i = log; i >= 0; i--)
        if(L[p] - (1 << i) >= L[q]) p = anc[p][i];
    if(p == q) return p;
    for(int i = log; i >= 0; 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] * 2;
}

int main()
{
    scanf("%d", &Q);
    n = 4;
    fa[2] = fa[3] = fa[4] = 1;
    L[2] = L[3] = L[4] = 1;
    anc[2][0] = anc[3][0] = anc[4][0] = 1;

    int U = 2, V = 3, diameter = 2;
    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
", diameter);
    }

    return 0;
}