Submission #1229726

#	Time	Username	Problem	Language	Result	Execution time	Memory
1229726		virgoliny	Beech Tree (IOI23_beechtree)	C++20	0 / 100	57 ms	19784 KiB

beechtree

#include "beechtree.h"
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#include<bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
typedef long long ll;
typedef long double ld;
#define endl "\n"
#define vll vector<ll>
#define sd second
#define ft first
#define all(x) x.begin(),x.end()
#define allr(x) x.rbegin(),x.rend()
#define pll pair<ll, ll>
#define mod 1000000007
#define _set tree<pll, null_type, less<pll>, rb_tree_tag, tree_order_statistics_node_update>
#define inf (ll)1e15
#define db(x) cout<<#x<<" : "<<x<<endl;
#define PRESICION(x) cout.setf(ios::fixed,ios::floatfield); cout.precision(x);
using namespace std;
using namespace __gnu_pbds;
ll dx[]={1, -1, 0, 0};
ll dy[]={0, 0, 1, -1};
inline ll sm(ll a, ll b){
return ((a%mod)+(b%mod))%mod;
}
inline ll ml(ll a, ll b){
return ((a%mod)*(b%mod))%mod;
}
inline ll rs(ll a, ll b){
return ((a%mod)-(b%mod)+mod)%mod;
}
ll bpow(ll a , ll b) {
if (b==0)return 1;
if (b%2!=0)return ((bpow(a, b-1)%mod)*(a%mod))%mod;
ll r=bpow (a ,b/ 2) ;
return ((r%mod)*(r%mod))%mod;
}
vector<vector<ll>> res, adj;/*
inline void dfs(ll n, ll p){
//db(n)
res[n].push_back(n);
for(auto y: adj[n]){
if(y==p)continue;
res.push_back()
}
for(auto y: adj[n]){
if(y==p)continue;
dfs(y, n);
for(auto x: res[y])res[n].push_back(x);
}
}
inline bool check(vector<int>& C, vector<int>& P, vector<ll>& v1){
vector<ll> v(501, 0);
bool n=1;
for(int i=1; i<v1.size(); i++){
if(v1[v[C[v1[i]]]]!=P[v1[i]])n=0;
v[C[v1[i]]]++;
}
return n;
}*/
std::vector<int> beechtree(int N, int M, std::vector<int> P, std::vector<int> C){
    vector<int> res1(N, 0);
    vector<bool> res2(N, 0);
    vector<vector<ll>>().swap(adj);
    vector<vector<ll>>().swap(res);
    adj.resize(N+1);
    res.resize(N+1);
    for(int i=1; i<N; i++){
    adj[P[i]].push_back(i);
    }
    set<ll> s, s1;
    for(int i=0; i<N; i++){
    set<ll>().swap(s);
    for(auto y: adj[i])s.insert(C[y]);
    ll o=s.size(), o1=adj[i].size();
    for(auto y: adj[i]){
    for(auto x: adj[y]){s.insert(C[x]);}
    }
    if(s.size()==o && o==o1)res2[i]=1;
    }
    for(int i=0; i<N; i++){
    ll o=0;
    bool n1=1;
    for(auto y: adj[i]){
    if(!adj[y].empty())o++;
    }
    if(o<=1 && res2[i]==1)res1[i]=1;
    }
    return res1;
}

• Subtask 0: 0 / 0
• Subtask 1: 0 / 9
• Subtask 2: 0 / 5
• Subtask 3: 0 / 9
• Subtask 4: 0 / 8
• Subtask 5: 0 / 14
• Subtask 6: 0 / 14
• Subtask 7: 0 / 12
• Subtask 8: 0 / 17
• Subtask 9: 0 / 12