Submission #1070087

#	Time	Username	Problem	Language	Result	Execution time	Memory
1070087		bleahbleah	Beech Tree (IOI23_beechtree)	C++17	26 / 100	2078 ms	136244 KiB

beechtree

#include "beechtree.h"
#include <bits/stdc++.h>
#define all(x) (x).begin(),(x).end()
using namespace std;

using ll = long long;
using ld = long double;

//#define int ll
#define sz(x) ((int)(x).size())

using pii = pair<int,int>;
using tii = tuple<int,int,int>;

const int nmax = 3e5 + 5;

const int mod = 998244853;
struct Mint {
  int val;
  Mint(ll x = 0): val((x % mod + mod) % mod) {;}
  Mint operator +(const Mint& x) const { return Mint(val + x.val); }
  Mint operator -(const Mint& x) const { return Mint(val - x.val); }
  Mint operator *(const Mint& x) const { return Mint((ll)val * x.val); }
  Mint operator +=(const Mint& x) { return *this = Mint(val + x.val); }
  Mint operator -=(const Mint& x) { return *this = Mint(val - x.val); }
  Mint operator *=(const Mint& x) { return *this = Mint((ll)val * x.val); }
  Mint operator ^(const int& _b) const {
    Mint accum = 1, a = *this;
    int b = _b;
    while(b) {
      accum = (b & 1? accum * a : accum);
      a *= a;
      b >>= 1;
    }
    return accum;
  }
  Mint operator /(const Mint& x) { return Mint((ll)val * (x ^ (mod - 2)).val); }
  Mint operator /=(const Mint& x) { return *this = Mint((ll)val * (x ^ (mod - 2)).val); }
};

Mint p[2][nmax];

#define hash bjsefdjhsdsfhoi
struct hash {
   Mint v[2];
   int len;
   hash(Mint a = 0, Mint b = 0, int c = 0) { v[0] = a; v[1] = b; len = c; }
   hash operator +(const hash& x) const {
      return hash(v[0] * p[0][x.len] + x.v[0], v[1] * p[1][x.len] + x.v[1], len + x.len);
   }
   hash operator -(const hash& x) const {
      return hash(v[0] - p[0][len - x.len] * x.v[0], v[1] - p[1][len - x.len] * x.v[1], len - x.len); 
   }
   bool operator !=(const hash& x) const {
      return v[0].val != x.v[0].val || v[1].val != x.v[1].val || len != x.len;
   }
   ll operator()() const { return (ll)v[0].val * mod + v[1].val; }
};



vector<pii> g[nmax];
vector<int> P, C;

unordered_set<int> colors[nmax], play[nmax];
bool isanc(unordered_set<int>& A, unordered_set<int>& B) {
   if(sz(A) < sz(B)) return 0;
   for(auto &x : B)
      if(!A.count(x)) return 0;
   return 1;
}

vector<int> sol;
int h[nmax];

int area[nmax];
hash subarb[nmax];

bool superimposable(int x, int y) { // x peste y 
   if(sz(g[x]) == 0 || x == y) return 1;
   if(subarb[x]() == subarb[y]()) return 1;
   //if(calc[x][y]) return SI[x][y];
   map<int, int> gy;
   for(auto [a, c] : g[y]) gy[c] = a;
   //calc[x][y] = 1;
   //SI[x][y] = 1; 
   for(auto [a, c] : g[x]) {
      if(gy.count(c) == 0) { return 0; }
      if(!superimposable(a, gy[c])) { return 0; }
   }
   return 1;
}

void insert(int& works, map<int, int>& st, int node) {
   if(!works) return;
   auto it = st.upper_bound(area[node]);
   if(it == st.end());
   else if(!superimposable(node, it -> second)) { works = 0; return; }
   if(it == st.begin()); 
   else if(!superimposable(prev(it) -> second, node)) { works = 0; return; }
   st[area[node]] = node;
   return;
}

map<int, int> dfs(int node) {
   map<int, int> here;
   sol[node] = 1;
   area[node] = 1;
   
   vector<hash> fils;
   
   for(auto [x, c] : g[node]) {
      auto T = dfs(x);
      fils.emplace_back(hash(c) + subarb[x]);
      area[node] += area[x];
      sol[node] &= sol[x];
      if(sz(T) > sz(here)) swap(here, T);
      for(auto [a, x] : T) insert(sol[node], here, x);
   }
   
   sort(all(fils), [&](auto a, auto b) { return a() < b(); });
   subarb[node] = hash(11);
   for(auto x : fils) subarb[node] = subarb[node] + x;
   subarb[node] = subarb[node] + hash(12);
   
   insert(sol[node], here, node);
   if(sz(g[node]) != sz(colors[node])) sol[node] = 0;
   return here;
}

std::vector<int> beechtree(int N, int M, std::vector<int> P_, std::vector<int> C_) {
   mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
   p[0][0] = p[1][0] = 1;
   p[0][1] = rng() % (mod - 1000) + 503;
   p[1][1] = rng() % (mod - 1200) + 505;
   for(int i = 2; i < nmax; i++)
      p[0][i] = p[0][i - 1] * p[0][1],
      p[1][i] = p[1][i - 1] * p[1][1];
      
   P = P_;
   C = C_;
   for(int i = 1; i < N; i++) {
      g[P[i]].emplace_back(i, C[i]);
      colors[P[i]].emplace(C[i]);
   }
   sol.assign(N, 0);
   dfs(0);
   
   
   return sol;
}

/**
      Töte es durch genaue Untersuchung\Töte es kann es nur noch schlimmer machen\Es lässt es irgendwie atmen
--
*/ 

• Subtask 0: 0 / 0
• Subtask 1: 9 / 9
• Subtask 2: 0 / 5
• Subtask 3: 9 / 9
• Subtask 4: 8 / 8
• Subtask 5: 0 / 14
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• Subtask 9: 0 / 12