Submission #428291

#	Time	Username	Problem	Language	Result	Execution time	Memory
428291		Blistering_Barnacles	Split the Attractions (IOI19_split)	C++14	7 / 100	105 ms	25896 KiB

split

#include "split.h"
//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, dp[105][100005], a[555555] , b[555555], k, l, m, n, o, p;
map < ll , ll > mp;
vll adj[555555] , v;
const ll mod = 1e9+7;
str s;
void dfs(ll u , ll par , ll cnt , ll op){
    //cout << cnt << endl ;
    if(a[u])R ;
    a[u] = op ;
    for(auto uu : adj[u]){
        if(uu == par)C ;
        dfs(uu , u , (cnt == b[op] ? 1 : cnt + 1) , (cnt == b[op] ? op + 1 : op)) ;
    }
}
vector<int> find_split(int n, int a1, int b1, int c, vector<int> p1, vector<int> q) {
    b[1] = a1 , b[2] = b1 , b[3] = c ,
    m = sz(p1) ;
    for(ll i = 1 ; i <= m ; i++){
        o = p1[i - 1] , p = q[i - 1] ;
        o++ , p++ ;
        adj[o].pb(p) , adj[p].pb(o) ;
    }
    o = 0 ;
    for(ll i = 1  ; i <= n ; i++){
        if(sz(adj[i]) == 1){
            o = i ;
            B ;
        }
    }
    if(!o)o = 1 ;
    dfs(o , 0 , 1 , 1) ;
    for(ll i = 1 ; i <= n ; i++)v.pb(a[i]) ;
    R v ;
}
/*
7 6 1 3 3
0 1
1 2
2 3
3 4
4 5
5 6
*/

• Subtask 1: 7 / 7
• Subtask 2: 0 / 11
• Subtask 3: 0 / 22
• Subtask 4: 0 / 24
• Subtask 5: 0 / 36