제출 #431285

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
431285		Blistering_Barnacles	슈퍼트리 잇기 (IOI20_supertrees)	C++14	56 / 100	279 ms	35212 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, par1[100005], a[555555] , par2[555555] , k, l, m, n, o, p;
map < ll , ll > mp;
const ll mod = 1e9+7;
str s;
vector < vll > ans ;
vll adj[555555] ;
ll getpar1(ll x){
    R (x == par1[x] ? x : par1[x] = getpar1(par1[x])) ;
}
void mrg1(ll x ,ll y){
    x = getpar1(x) , y = getpar1(y) ;
    if(x == y)R ;
    par1[y] = x ;
}
ll getpar2(ll x){
    R (x == par2[x] ? x : par2[x] = getpar2(par2[x])) ;
}
void mrg2(ll x , ll y){
    x = getpar2(x) , y = getpar2(y) ;
    if(x == y)R ;
    par2[y] = x ;
}
int construct(vector<vector<int>> p) {
    n = sz(p) ;
    ans.resize(n) ;
    for(ll i = 0 ; i < n ; i++){
        if(p[i][i] != 1)R 0 ;
        ans[i].resize(n) ;
        par1[i] = par2[i] = i ;
    }
    for(ll i = 0 ; i < n ; i++){
        for(ll j = i + 1 ; j < n ; j++){
            if(p[i][j] == 1)mrg1(i , j) ;
            else if(p[i][j] == 2)mrg2(i , j) ;
        }
    }
    for(ll i = 0 ; i < n ; i++){
        if(par1[i] == i){
            ll op = getpar2(i) ;
            adj[op].pb(i) ;
            C ;
        }
        ll op = getpar1(i) ;
        ans[op][i] = ans[i][op] = 1 ;
    }
    for(ll i = 0 ; i < n ; i++){
        m = sz(adj[i]) ;
        if(m <= 1)C ;
        if(m == 2)R 0 ;
        for(ll j = 0 ; j < m ; j++){
            ll j1 = adj[i][j] , j2 = adj[i][(j + 1) % m] ;
            //cout << j1 << " " << j2 << endl ;
            ans[j1][j2] = ans[j2][j1] = 1 ;
        }
    }
    for(ll i = 0; i < n ; i++){
        for(ll j = i + 1 ; j < n ; j++){
            ll o1 = getpar1(i) , o2 = getpar1(j) ;
            if(o1 == o2 && p[i][j] != 1)R 0 ;
            /*o1 = getpar2(i) , o2 = getpar2(j) ;
            if(o1 == o2 && p[i][j] != 2)R 0 ;*/

        }
    }
    for(ll i = 0 ; i < n ; i++)ans[i][i] = 0 ;
    build(ans) ;
    R 1 ;
}

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