제출 #1309942

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1309942		Tymond	Bitaro’s Party (JOI18_bitaro)	C++20	14 / 100	447 ms	220484 KiB

bitaro

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
#define fi first
#define se second
#define vi vector<int>
#define vll vector<long long>
#define pii pair<int, int>
#define pll pair<long long, long long>
#define pb push_back
#define mp make_pair
#define eb emplace_back
#define all(x) (x).begin(), (x).end()
#define sz(x) (int)(x).size()
mt19937 rng(chrono::high_resolution_clock::now().time_since_epoch().count());
mt19937_64 rng64(chrono::high_resolution_clock::now().time_since_epoch().count());
inline int rand(int l,int r){return uniform_int_distribution<int>(l, r)(rng);}
inline ll rand(ll l,ll r){return uniform_int_distribution<ll>(l, r)(rng64);}
#ifdef DEBUG
auto&operator<<(auto&o,pair<auto,auto>p){return o<<"("<<p.first<<", "<<p.second<<")";}
auto operator<<(auto&o,auto x)->decltype(x.end(),o){o<<"{";int i=0;for(auto e:x)o<<","+!i++<<e;return o<<"}";}
#define debug(X...)cerr<<"["#X"]: ",[](auto...$){((cerr<<$<<"; "),...)<<endl;}(X)
#else
#define debug(...){}
#endif

struct custom_hash {
    static uint64_t splitmix64(uint64_t x) {
        x += 0x9e3779b97f4a7c15;
        x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
        x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
        return x ^ (x >> 31);
    }

    size_t operator()(uint64_t x) const {
        static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
        return splitmix64(x + FIXED_RANDOM);
    }
};

struct pair_hash{
  size_t operator()(const pair<int,int>&x)const{
    return hash<long long>()(((long long)x.first)^(((long long)x.second)<<32));
  }
};

const int INF = 1e9;
const int MAXN = 1e5 + 7;
const int B = 267;
vi g[MAXN];
vector<pii> best[MAXN];
int blocked[MAXN];
int dist[MAXN];
int n, m, q;

void preprocessing(){
    for(int i = 1; i <= n; i++){
        best[i].pb(mp(0, i));
        for(auto u : g[i]){
            vector<pii> merged = {};
            int wsk1 = 0;
            int wsk2 = 0;
            while(sz(merged) <= B && (wsk1 < sz(best[i]) || wsk2 < sz(best[u]))){
                if(wsk1 == sz(best[i])){
                    merged.pb(mp(best[u][wsk2].fi + 1, best[u][wsk2].se));
                    wsk2++;
                }else if(wsk2 == sz(best[u])){
                    merged.pb(mp(best[i][wsk1].fi, best[i][wsk1].se));
                    wsk1++;
                }else{
                    if(best[u][wsk2].fi + 1 >= best[i][wsk1].fi){
                        merged.pb(mp(best[u][wsk2].fi + 1, best[u][wsk2].se));
                        wsk2++;
                    }else{
                        merged.pb(mp(best[i][wsk1].fi, best[i][wsk1].se));
                        wsk1++;
                    }
                }
            }
            best[i] = merged;
        }
        //debug(best[i]);
    }
}

int solveBig(int v){
    for(int i = 1; i <= v; i++){
        if(blocked[i]){
            dist[i] = -INF;
        }else{
            dist[i] = 0;
        }

        for(auto u : g[i]){
            if(dist[u] != -INF){
                dist[i] = max(dist[i], dist[u] + 1);
            }
        }
    }

    if(dist[v] == -INF){
        return -1;
    }
    return dist[v];
}

int main(){
    ios_base::sync_with_stdio(0);
    cin.tie(NULL);
    cout.tie(NULL);

    cin >> n >> m >> q;
    for(int i = 1; i <= m; i++){
        int a, b;
        cin >> a >> b;
        g[b].pb(a);
    }

    preprocessing();
    while(q--){
        int v, l;
        cin >> v >> l;
        vi curr(l);
        for(int i = 0; i < l; i++){
            cin >> curr[i];
            blocked[curr[i]] = 1;
        }

        if(l >= B){
            cout << solveBig(v) << '\n';
        }else{
            int ans = -1;
            for(auto ele : best[v]){
                if(!blocked[ele.se]){
                    ans = max(ans, ele.fi);
                }
            }
            cout << ans << '\n';
        }
        for(auto ele : curr){
            blocked[ele] = 0;
        }
    }

    return 0;
}

• Subtask 1: 7 / 7
• Subtask 2: 7 / 7
• Subtask 3: 0 / 86