牛客多校第四场 Ancient Distance

考虑本题如何暴力：

可以枚举用了多少个关键点，在确定关键点个数之后，就可以二分最小距离x了，至于二分的check是一个贪心的过程，显然，每次选择最深的那个点，然后向上找他的x级祖先，然后把祖先的子树删掉就好了。这样就可以暴力出来了

换个思路，我们设F(x)为最小距离为x时需要多少关键点，Ans[F(x)] = x 的意义就是，当花费F(x)个关键点的时候，最小距离为x，那么对Ans求和就好了。当然Ans[F(x)]需要求前缀min，为什么呢？有一个感性的理解，用的关键点越多，那么必然距离会有一个小于等于的关系。还有F(x)不一定覆盖所有的1-n,但是因为存在小于等于关系，所以可以全部求出。

//#define LOCAL
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define mem(a, b) memset(a,b,sizeof(a))
#define sz(a) (int)a.size()
#define INF 0x3f3f3f3f
#define DNF 0x7f
#define DBG printf("this is a input\n")
#define fi first
#define se second
#define mk(a, b) make_pair(a,b)
#define pb push_back
#define LF putchar('\n')
#define SP putchar(' ')
#define p_queue priority_queue
#define CLOSE ios::sync_with_stdio(0); cin.tie(0)

template<typename T>
void read(T &x) {x = 0;char ch = getchar();ll f = 1;while(!isdigit(ch)){if(ch == '-')f *= -1;ch = getchar();}while(isdigit(ch)){x = x * 10 + ch - 48; ch = getchar();}x *= f;}
template<typename T, typename... Args>
void read(T &first, Args& ... args) {read(first);read(args...);}
template<typename T>
void write(T arg) {T x = arg;if(x < 0) {putchar('-'); x =- x;}if(x > 9) {write(x / 10);}putchar(x % 10 + '0');}
template<typename T, typename ... Ts>
void write(T arg, Ts ... args) {write(arg);if(sizeof...(args) != 0) {putchar(' ');write(args ...);}}
using namespace std;

ll gcd(ll a, ll b) {
    return b == 0 ? a : gcd(b, a % b);
}

ll lcm(ll a, ll b) {
    return a / gcd(a, b) * b;
}
const int N = 2e5+10;

int n;
struct node
{
    int t, w, next;
}edge[N << 1];
vector <int> ve;
int head[N], dfn[N], fdfn[N], siz[N], dep[N], cnt, times = 0;
int bz[N][25];
void add (int f, int t, int w)
{
    edge[cnt].t = t;
    edge[cnt].w = w;
    edge[cnt].next = head[f];
    head[f] = cnt ++;
}
void dfs (int u , int fa)
{
    dfn[u] = ++ times;
    fdfn[times] = u;
    siz[u] = 1;
    for (int i = head[u] ; i != -1 ; i = edge[i].next)
    {
        int v = edge[i].t;
        if (v != fa)
        {
            bz[v][0] = u;
            dep[v] = dep[u] + 1;
            dfs (v , u);
            siz[u] += siz[v];
        }
    }
}
void deal()
{
    for (int i = 1 ; i <= 20 ; i ++)
        for (int j = 1 ; j <= n ; j ++)
            bz[j][i] = bz[bz[j][i-1]][i-1];
}
int val1[N << 2] , val2[N << 2], lz[N << 2], ans[N];
void push_up(int rt)
{
    if (dep[val1[rt<<1]] > dep[val1[rt<<1|1]])
        val1[rt] = val1[rt<<1];
    else
        val1[rt] = val1[rt<<1|1];
}
void pushdown(int rt)
{
    if (lz[rt])
    {
        ve.pb(rt << 1);
        ve.pb(rt << 1 | 1);
        val1[rt << 1] = val1[rt << 1 | 1] = 0;
        lz[rt << 1] = lz[rt << 1 | 1] = 1;
        lz[rt] = 0;
    }
}
void build (int rt, int l ,int r)
{
    lz[rt] = 0;
    if (l == r)
    {
        val1[rt] = val2[rt] = fdfn[l];
        return ;
    }
    int mid = (l + r) >> 1;
    build (rt << 1 , l , mid);
    build (rt << 1 | 1 , mid + 1 , r);
    push_up(rt);
    val2[rt] = val1[rt];
}
int kfa (int u , int k)
{
    for(int i = 20 ; i >= 0 ; i --)
        if(k >= (1 << i))
            k -= (1 << i) , u = bz[u][i];
    return u ? u : 1;
}
void update(int rt, int L, int R, int l, int r)
{
    ve.pb(rt);
    if(l > R || r < L)
        return;
    if(l >= L && r <= R) {
        val1[rt] = 0, lz[rt] = 1;
        return;
    }
    pushdown(rt);
    int mid = (l + r) >> 1;
    update(rt << 1 , L, R, l, mid);
    update(rt << 1 | 1, L, R, mid+1, r);
    push_up(rt);
}
int get (int x)
{
    ve.clear();
    int ret = 0 , now , fnow;
    while(true)
    {
        now = val1[1];
        if (now == 0) break;
        fnow = kfa (now, x);
        update (1, dfn[fnow], dfn[fnow] + siz[fnow] - 1, 1 , n);
        ++ ret;
    }
    for (auto i : ve)
        val1[i] = val2[i], lz[i] = 0;
    return ret;
}
void init ()
{
    times = cnt = 0;
    for (int i = 1 ; i <= n ; i ++)
        dep[i] = 0, fdfn[i] = 0 , siz[i] = 0, dfn[i] = 0, ans[i] = n + 1, head[i] = -1;
}
int main()
{
    while (scanf("%d",&n) != EOF)
    {
        init();
        for (int i = 2 ; i <= n ; i ++)
        {
            int u;
            read (u);
            add (i , u, 1);
            add (u , i, 1);
        }
        dep[1] = 1;
        dfs (1, 1);
        deal ();
        build (1, 1, n);
        //get(i) :最近距离为i时需要get(i)个关键点
        for (int i = n ; i >= 0 ; i --) {
            ans[get(i)] = i;
            //cout << get(i) << ' ' << i << endl;
        }
        for (int i = 2 ; i <= n ; i ++)
            ans[i] = min (ans[i-1], ans[i]);
        ll ret = 0;
        for (int i = 1 ; i <= n ; i ++)
            ret += ans[i];
        write(ret), LF;

    }
}
/*
5
1 1 3 3
 */