Submission #955208

#	Time	Username	Problem	Language	Result	Execution time	Memory
955208		amine_aroua	Railway Trip 2 (JOI22_ho_t4)	C++17	11 / 100	2053 ms	16896 KiB

Main

#include <bits/stdc++.h>
//#pragma GCC optimize("O3")
//#pragma GCC optimize("unroll-loops")
using namespace std;
#define int long long
#define vi vector<int>
#define vl vector<long long>
#define vii vector<pair<int,int>>
#define vll vector<pair<long long,long long>>
#define pb push_back
#define ll long long
#define ld long double
#define nl '\n'
#define boost ios::sync_with_stdio(false)
#define mp make_pair
#define se second
#define fi first
#define fore(i, y) for(int i = 0; i < y; i++)
#define forr(i,x,y) for(int i = x;i<=y;i++)
#define forn(i,y,x) for(int i = y; i >= x; i--)
#define all(v) v.begin(),v.end()
#define sz(v) (int)v.size()
#define clr(v,k) memset(v,k,sizeof(v))
#define rall(v) v.rbegin() , v.rend()
#define pii pair<int,int>
#define pll pair<ll , ll>

const ll MOD = 1e9 + 7;
const ll INF = 1e18 + 1;

ll gcd(ll a , ll b) {return b ? gcd(b , a % b) : a ;} // greatest common divisor (gcd)
ll lcm(ll a , ll b) {return a * (b / gcd(a , b));} // least common multiple (lcm)

// HERE IS THE SOLUTION
vi mxTree , mnTree;
int n;
void build()
{
    while(__builtin_popcount(n) != 1)
        n++;
    mxTree.assign(2*n , 0);
    mnTree.assign(2*n , n + 1);
}
void upd(int node , int L , int R , int l , int  r , int x , bool type)
{
    if(l <= L && R <= r)
    {
        if(!type)
            mxTree[node] = max(mxTree[node] , x);
        else
            mnTree[node] = min(mnTree[node] , x);
        return ;
    }
    if(l > R || L > r)
        return ;
    int mid = (L + R)/2;
    upd(2*node , L , mid , l , r , x ,  type);
    upd(2*node + 1 , mid + 1 , R , l , r , x , type);
}
int getMax(int i)
{
    int ans = i;
    for(int j = (i + n); j>= 1;j/=2)
    {
        ans = max(ans , mxTree[j]);
    }
    return ans;
}
int getMin(int i)
{
    int ans = i;
    for(int j = (i + n); j>= 1;j/=2)
    {
        ans = min(ans , mnTree[j]);
    }
    return ans;
}

signed main()
{
    boost;
    cin.tie(0);
    cout.tie(0);
    int N , K , M;
    cin>>N>>K>>M;
    n = N +1;
    build();
    vector<vi> adj(N + 1);
    forr(i , 1  , N)
    {
        adj[i] = {i , i};
    }
    fore(i , M)
    {
        int A , B;
        cin>>A>>B;
        if(A < B)
        {
            upd(1 , 0 , n - 1 , A , min(B - 1 , A + K - 1) , B , 0);
        }
        else
        {
            upd(1 , 0 , n - 1 , max(B + 1 , A - K + 1) , A , B , 1);
        }
    }
    forr(i , 1 , N)
    {
        adj[i][0] = getMin(i);
        adj[i][1] = getMax(i);
    }
    int Q;
    cin>>Q;
    while(Q--)
    {
        int S , T;
        cin>>S>>T;
        queue<int> q;
        q.push(S);
        vi dist(N + 1 , INF);
        set<int> free;
        forr(i , 1 , N)
            free.insert(i);
        dist[S] = 0;
        while(!q.empty())
        {
            int cur = q.front();
            q.pop();
            if(!free.count(cur))
                continue;
            free.erase(cur);
            if(cur == T)
                break;
            int mn = INF , mx = -INF;
            int mnL = -1, mxR = -1;
            auto it1 = free.lower_bound(adj[cur][0]);
            auto it2 = free.upper_bound(adj[cur][1]);
            for(auto it = it1 ; it!= it2; it++)
            {
                int u = *it;
                if(dist[u] > dist[cur] + 1)
                    dist[u] = dist[cur] + 1;
                if(adj[u][0] < mn)
                {
                    mn = adj[u][0];
                    mnL = u;
                }
                if(adj[u][1] > mx)
                {
                    mx = adj[u][1];
                    mxR = u;
                }
            }
            vector<int> to_erase;
            for(auto it = it1 ; it!= it2; it++)
            {
                int u = *it;
                if(u == mnL || u == mxR)
                    continue;
                to_erase.pb(u);
            }
            for(auto it : to_erase)
                free.erase(it);
            if(mnL != -1)
                q.push(mnL);
            if(mxR != -1)
                q.push(mxR);
        }
        cout<<(dist[T] == INF ? -1 : dist[T])<<nl;
    }
}

• Subtask 1: 0 / 8
• Subtask 2: 0 / 8
• Subtask 3: 11 / 11
• Subtask 4: 0 / 25
• Subtask 5: 0 / 35
• Subtask 6: 0 / 13