제출 #425822

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
425822		Blistering_Barnacles	장난감 기차 (IOI17_train)	C++11	10 / 100	2076 ms	14260 KiB

train

#include "train.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[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;
vi o1 , o2 ;
vll cur ;
ll vis[555555] , nxt[555555] ;
vll charging , whom;
vi ans ;
ll mem(ll i , ll cnt){
    if(charging[i])cnt = n ;
    if(!cnt)R 0 ;
    if(nxt[i] != -1){
        if(charging[i])R 1 ;
        R mem(nxt[i] , cnt - 1) ;
    }
    ll ret = !whom[i] ;
    for(auto u : adj[i]){
        nxt[i] = u ;
        ll op = mem(u , cnt - 1) ;
        if(!whom[i])ret &= op ;
        else ret |= op ;
        nxt[i] = -1 ;
    }
    R ret ;
}
vector<int> who_wins(std::vector<int> who, std::vector<int> charge, std::vector<int> u, std::vector<int> v) {
    n = sz(who) , m = sz(u) ;
    for(ll i = 0 ; i < m ; i++){
        adj[u[i]].pb(v[i]) ;
    }
    charging = charge ;
    whom = who ;
    for(ll i = 0 ; i < n ; i++){
        for(ll i = 0 ; i < n ; i++)nxt[i] = -1 ;
        ans.pb(mem(i , n)) ;
    }
    R ans ;
}

• Subtask 1: 0 / 5
• Subtask 2: 10 / 10
• Subtask 3: 0 / 11
• Subtask 4: 0 / 11
• Subtask 5: 0 / 12
• Subtask 6: 0 / 51