oj.uz

Submission #982837

# Submission time Handle Problem Language Result Execution time Memory
982837 2024-05-14T19:29:21 Z vjudge1 Game (APIO22_game) C++17
0 / 100
 3 ms 16728 KB 
#include <bits/stdc++.h>
#define endl '\n'
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define fo(i,n) for(auto i =0 ; i < n;i++)
#define fore(i,l,r) for(auto i = l; i < r;i++)
#define forex(i,r,l) for(auto i = r; i >= l; i--)
#define ffo(i,n) forex(i,n-1,0)
#define all(x) x.begin(),x.end()
#define lsb(x) x&(-x)
#define sz(x) (int)x.size()
#define gcd(a,b) __gcd(a,b)
#define vii vector<ii>
using namespace std;
using ll = long long; using ull = unsigned long long;
using vi = vector<int>;using ii = pair<int,int>;using mii = map<int,int>;
// #pragma GCC target ("avx2")
// #pragma GCC optimization ("O3")
// #pragma GCC optimization ("unroll-loops")
const int N = 3e5 + 5;
int n,k;
int vis[N], timer = 0;
vi graph[N], graphinv[N];
// mn[i] la minima a la que puede llegar el nodo i
// mx[i] la maxima <= k que puede llegar al nodo i 
int mx[N], mn[N];
void init(int N, int K){
    n=N,k=K;
    fo(i, n){
        mx[i] = -1;
        mn[i] = 1e9;
        // i no puede llegar a i, a menos que hagas edge(i,i)
        // entonces la minima es i+1 y max i-1
        if(i>0 and k>1){mx[i]=i-1;graphinv[i].pb(i-1);}
        if(i<k-1){mn[i]=i+1;graph[i].pb(i+1);}
    }
}
bool ya = 0;
int add_teleporter(int u, int v){
    if(mn[v]<=mx[u])ya=1;
    if(ya) return 1;
    graph[u].pb(v);graphinv[v].pb(u);
    queue<int> q;timer++;
    q.push(u);
    while(!q.empty()){
        int nodo = q.front();q.pop();
        if(ya) return 1;
        mn[nodo] = min(mn[nodo], mn[v]);
        if(mn[nodo]<=mx[nodo]){ya=1;return 1;}
        for(int v : graphinv[nodo]){
            if(vis[v] == timer) continue;
            vis[v] = timer;
            q.push(v);
        }
    }timer++;
    q.push(v);
    while(!q.empty()){
        int nodo = q.front(); q.pop();
        if(ya) return 1;
        mx[nodo] = max(mx[nodo] , mx[u]);
        if(mn[nodo] <= mx[nodo]){ya = 1; return 1;}
        for(int v : graph[nodo]){
            if(vis[v] == timer)continue;
            vis[v] = timer;
            q.push(v);
        }
    }
    if(mn[v]<=mx[u])ya=1;
    fo(i, k) if(mn[i] <= mx[i] ){ya =1; break;}
    if(ya)return 1;
    return 0;
}

Subtask #1
0.0 / 2.0

# Verdict Execution time Memory Grader output
1 Incorrect 3 ms 16728 KB Wrong Answer[1]
2 Halted 0 ms 0 KB -

Subtask #2
0.0 / 10.0

# Verdict Execution time Memory Grader output
1 Incorrect 3 ms 16728 KB Wrong Answer[1]
2 Halted 0 ms 0 KB -

Subtask #3
0.0 / 18.0

# Verdict Execution time Memory Grader output
1 Incorrect 3 ms 16728 KB Wrong Answer[1]
2 Halted 0 ms 0 KB -

Subtask #4
0.0 / 30.0

# Verdict Execution time Memory Grader output
1 Incorrect 3 ms 16728 KB Wrong Answer[1]
2 Halted 0 ms 0 KB -

Subtask #5
0.0 / 40.0

# Verdict Execution time Memory Grader output
1 Incorrect 3 ms 16728 KB Wrong Answer[1]
2 Halted 0 ms 0 KB -