题目大致意思就是给出含有n节点的树，1为树根。从1开始bfs搜索整棵树，不过有顺序。深度小的比大的优先，
同一深度的其优先值为其父节点在上一层的搜索顺序，如果同属一个父节点，那么就是节点编号小的优先。然后求出路径总和sum。比如这次搜索到了u，下次需要搜索到v，
那么sum += dis(u, v)。树上的dis就是lca了，因为边权为1.
然后剩下的就是模拟这个过程求解了。
/*****************************************
Author :Crazy_AC(JamesQi)
Time :2016
File Name :
*****************************************/
// #pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <sstream>
#include <string>
#include <stack>
#include <queue>
#include <deque>
#include <vector>
#include <map>
#include <set>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <climits>
using namespace std;
#define MEM(x,y) memset(x, y,sizeof x)
#define pk push_back
#define lson rt << 1
#define rson rt << 1 | 1
#define bug cout << "BUG HERE\n"
#define debug(x) cout << #x << " = " << x << endl
#define ALL(v) (v).begin(), (v).end()
#define lowbit(x) ((x)&(-x))
#define Unique(x) sort(ALL(x)); (x).resize(unique(ALL(x)) - (x).begin())
#define BitOne(x) __builtin_popcount(x)
#define showtime printf("time = %.15f\n",clock() / (double)CLOCKS_PER_SEC)
#define Rep(i, l, r) for (int i = l;i <= r;++i)
#define Rrep(i, r, l) for (int i = r;i >= l;--i)
typedef long long LL;
typedef unsigned long long ULL;
const int maxn = 1e5 + 123;
const int LOGN = 20;
int head[maxn], nxt[maxn*2], pnt[maxn*2], ecnt;
inline void addedge(int u,int v) {
pnt[ecnt] = v, nxt[ecnt] = head[u], head[u] = ecnt++;
pnt[ecnt] = u, nxt[ecnt] = head[v], head[v] = ecnt++;
}
// vector<vector<int> > G;
int n;
int fa[maxn];
struct node {
int pre, u;
//pre为其父节点的优先值,pre相同时就是u小的优先
//同一层访问的优先顺序
bool operator < (const node& rhs) const {
return pre < rhs.pre || (pre == rhs.pre && u < rhs.u);
}
};
struct _LCA {
int dep[maxn];
int fa[LOGN][maxn];
void dfs(int u,int pre, int depth) {
fa[0][u] = pre;dep[u] = depth;
for (int i = head[u];~i;i = nxt[i]) {
if (pnt[i] == pre) continue;
dfs(pnt[i], u, depth + 1);
}
}
inline void Build(int n) {
for (int k = 0;k < LOGN - 1;++k) {
for (int u = 1;u <= n;++u) {
if (fa[k][u] == -1) fa[k + 1][u] = -1;
else fa[k + 1][u] = fa[k][ fa[k][u] ];
}
}
}
inline int upslope(int u,int p) {
for (int k = 0;k < LOGN - 1;++k)
if ((p>>k) & 1) u = fa[k][u];
return u;
}
inline int LCA(int u,int v) {
if (dep[u] < dep[v]) swap(u, v);
u = upslope(u, dep[u] - dep[v]);
if (u == v) return u;
for (int k = LOGN - 1;k >= 0;--k) {
if (fa[k][u] != fa[k][v])
u = fa[k][u], v = fa[k][v];
}
return fa[0][u];
}
LL get_dis(int u, int v) {
int lca = LCA(u, v);
return (LL)dep[u] + (LL)dep[v] - 2LL * (LL)dep[lca];
}
}lca;
inline void bfs_init() {
LL sum = 0;
vector<node> vec;
for (int i = head[1];~i;i = nxt[i]) {
int v = pnt[i];
vec.push_back(node{
1, v});
}
int last = 1;//上一层访问的最后一个点的编号
while(!vec.empty()) {
vec.push_back(node{-1, last});
sort(ALL(vec));
int size = vec.size();
for (int i = 1;i < size;++i) {
sum += (LL)lca.get_dis(vec[i].u, vec[i - 1].u);
}
vector<node> temp;
Rep(i, 1, size - 1) {
int u = vec[i].u;
for (int j = head[u];~j;j = nxt[j]) {
int v = pnt[j];
if (v == fa[u]) continue;
temp.push_back(node{i + 1, v});
}
}
last = vec.back().u;
vec.clear();
vec = temp;
}
cout << sum << endl;
}
int main(int argc, const char * argv[])
{
// freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout);
// ios::sync_with_stdio(false);
// cout.sync_with_stdio(false);
// cin.sync_with_stdio(false);
while(~scanf("%d", &n)) {
// G.clear();
// G.resize(n + 2);
memset(head, -1, sizeof head), ecnt = 0;
Rep(i, 2, n) {
scanf("%d", &fa[i]);
addedge(i, fa[i]);
}
lca.dfs(1, -1, 0);
lca.Build(n);
bfs_init();
}
// showtime;
return 0;
}