2020/finals/somebody_elses_problem.cpp

// Somebody Else's Problem
// Solution by Jacob Plachta

#include <algorithm>
#include <functional>
#include <numeric>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <complex>
#include <cstdlib>
#include <ctime>
#include <cstring>
#include <cassert>
#include <string>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <unordered_map>
#include <unordered_set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <sstream>
using namespace std;

#define LL long long
#define LD long double
#define PR pair<int,int>

#define Fox(i,n) for (i=0; i<n; i++)
#define Fox1(i,n) for (i=1; i<=n; i++)
#define FoxI(i,a,b) for (i=a; i<=b; i++)
#define FoxR(i,n) for (i=(n)-1; i>=0; i--)
#define FoxR1(i,n) for (i=n; i>0; i--)
#define FoxRI(i,a,b) for (i=b; i>=a; i--)
#define Foxen(i,s) for (auto i:s)
#define Min(a,b) a=min(a,b)
#define Max(a,b) a=max(a,b)
#define Sz(s) int((s).size())
#define All(s) (s).begin(),(s).end()
#define Fill(s,v) memset(s,v,sizeof(s))
#define pb push_back
#define mp make_pair
#define x first
#define y second

template<typename T> T Abs(T x) { return(x < 0 ? -x : x); }
template<typename T> T Sqr(T x) { return(x * x); }
string plural(string s) { return(Sz(s) && s[Sz(s) - 1] == 'x' ? s + "en" : s + "s"); }

const int INF = (int)1e9;
const LD EPS = 1e-12;
const LD PI = acos(-1.0);

#define GETCHAR getchar_unlocked

bool Read(int& x) {
  char c, r = 0, n = 0;
  x = 0;
  for (;;) {
    c = GETCHAR();
    if ((c < 0) && (!r))
      return(0);
    if ((c == '-') && (!r))
      n = 1;
    else if ((c >= '0') && (c <= '9'))
      x = x * 10 + c - '0', r = 1;
    else if (r)
      break;
  }
  if (n)
    x = -x;
  return(1);
}

#define MOD 1000000007
#define LIM 1000001

int N, ans;
vector<int> ch[LIM];
int dyn1[LIM]; // max. len. of downward i -> leaf
int dyn2[LIM]; // max. len. of downward root -> leaf outside i's subtree
int dyn3[LIM]; // max. disjoint len. of downward root -> leaf plus other path ongoing from i's parent

void rec1(int i)
{
  // recurse, and compute dyn1[i]
  dyn1[i] = 0;
  Foxen(c, ch[i])
  {
    rec1(c);
    Max(dyn1[i], dyn1[c] + 1);
  }
}

void rec2(int i, int d)
{
  // compute answer for i
  ans = ans * (LL)max(dyn1[i] + dyn2[i] + (!i ? 0 : 1), dyn3[i]) % MOD;
  // compute max 2 (dyn1[c] + 1, c) pairs
  PR m1, m2;
  m1 = m2 = mp(0, -1);
  Foxen(c, ch[i])
  {
    PR p = mp(dyn1[c] + 1, c);
    if (p > m1)
      m2 = m1, m1 = p;
    else if (p > m2)
      m2 = p;
  }
  // compute dyn2[c] and dyn3[c], and recurse
  Foxen(c, ch[i])
  {
    int m = (m1.y == c ? m2 : m1).x;
    dyn2[c] = max(d + m, dyn2[i]);
    dyn3[c] = max(dyn3[i] + 1, dyn2[i] + m + 1 + (!i ? 0 : 1));
    rec2(c, d + 1);
  }
}

int ProcessCase()
{
  int i, j;
  // input
  Read(N);
  Fox1(i, N - 1)
  {
    Read(j), j--;
    ch[j].pb(i);
  }
  // DP
  ans = 1;
  rec1(0);
  rec2(0, 0);
  // reset
  Fox(i, N)
    ch[i].clear();
  return(ans);
}

int main()
{
  int T, t;
  Read(T);
  Fox1(t, T)
    printf("Case #%d: %d\n", t, ProcessCase());
  return(0);
}