Submission #431035

#	Time	Username	Problem	Language	Result	Execution time	Memory
431035		Blistering_Barnacles	Connecting Supertrees (IOI20_supertrees)	C++14	0 / 100	10 ms	13348 KiB

supertrees

#include "supertrees.h"
#include <vector>
//apig's property
//Happiness can be found, even in the darkest of times, if one only remembers to turn on the light
//El Pueblo Unido Jamas Sera Vencido
//The saddest thing about betrayal? is that it never comes from your enemies
//Do or do not... there is no try
//Billions of bilious blue blistering barnacles in a thundering typhoon!
#include<bits/stdc++.h>
#define fast ios_base::sync_with_stdio(0) , cin.tie(0) , cout.tie(0)
#define F first
#define S second
#define pb push_back
#define vll vector< ll >
#define vi vector< int >
#define pll pair< ll , ll >
#define pi pair< int , int >
#define all(s) s.begin() , s.end()
#define sz(s) s.size()
#define str string
#define md ((s + e) / 2)
#define mid ((l + r) / 2)
#define msdp(dp) memset(dp , -1 , sizeof dp)
#define mscl(dp) memset(dp , 0 , sizeof dp)
#define C continue
#define R return
#define B break
#define lx node * 2
#define rx node * 2 + 1
#define br(o) o ; break
#define co(o) o ; continue
using namespace std;
typedef int ll;
ll q, par[100005], a[555555] , b[555555], k, l, m, n, o, p;
map < ll , ll > mp;
vll adj[555555];
const ll mod = 1e9+7;
str s;
vector < vll > ans ;
ll getpar(ll x){
    R (x == par[x] ? x : par[x] = getpar(par[x])) ;
}
void mrg(ll x , ll y){
    x = par[x] , y = par[y] ;
    if(x == y)R ;
    if(sz(adj[y]) > sz(adj[x]))swap(x , y) ;
    for(auto u : adj[y]){
        adj[x].pb(u) ;
    }
    adj[y].clear() ;
    par[y] = x ;
}
int construct(vector<vector<int>> p) {
    n = sz(p) ;
    ans.resize(n) ;
    for(ll i = 0 ; i < n ; i++){
        par[i] = i ;
        adj[i].pb(i) ;
    }
    for(ll i = 0 ; i < n ; i++){
        if(p[i][i] != 1)R 0 ;
        ans[i].resize(n) ;
        for(ll j = i + 1 ; j < n ; j++){
            if(p[i][j] == 2)mrg(i , j) ;
        }
    }
    for(ll i = 0 ; i < n ; i++){
        for(ll j = 0 ; j < n ; j++){
            if(i == j)C ;
            ll o1 = getpar(i) , o2 = getpar(j) ;
            if(o1 == o2 && !p[i][j])R 0 ;
        }
    }
    for(ll i = 0 ; i < n ; i++){
        if(getpar(i) != i)C ;
        if(sz(adj[i]) == 1)C ;
        if(sz(adj[i]) == 2)C ; 
        m = sz(adj[i]) ;
        for(ll j = 0 ; j < m ; j++){
            ll j1 = adj[i][j] ;
            ll j2 = adj[i][(j + 1) % m] ;
            ans[j1][j2] = ans[j2][j1] = 1 ;
        }
    }
    for(ll i = 0 ; i < n ; i++)ans[i][i] = 0 ;
    build(ans) ;
    R 1 ;
}

• Subtask 1: 0 / 11
• Subtask 2: 0 / 10
• Subtask 3: 0 / 19
• Subtask 4: 0 / 35
• Subtask 5: 0 / 21
• Subtask 6: 0 / 4