Submission #674980

#	Time	Username	Problem	Language	Result	Execution time	Memory
674980		urosk	Treatment Project (JOI20_treatment)	C++14	39 / 100	2300 ms	524288 KiB

treatment

#define here cerr<<"===========================================\n"
#define dbg(x) cerr<<#x<<": "<<x<<endl;
#include "bits/stdc++.h"
//#include <ext/pb_ds/tree_policy.hpp>
//#include <ext/pb_ds/assoc_container.hpp>
#define ld double
#define ll long long
#define llinf 100000000000000000LL // 10^17
#define pb push_back
#define popb pop_back
#define fi first
#define sc second
#define endl '\n'
#define pll pair<ll,ll>
#define pld pair<ld,ld>
#define sz(a) (ll)(a.size())
#define all(a) a.begin(),a.end()
#define ceri(a,l,r) {cerr<<#a<<": ";for(ll i_ = l;i_<=r;i_++) cerr<<a[i_]<< " ";cerr<<endl;}
#define cer(a) {cerr<<#a<<": ";for(ll x_ : a) cerr<<x_<< " ";cerr<<endl;}

#define daj_mi_malo_vremena ios_base::sync_with_stdio(false);cerr.tie(0);cout.tie(0);cin.tie(0);

using namespace std;
//using namespace __gnu_pbds;
/*
ll add(ll x,ll y){
    x+=y;
    if(x<0){
        x%=mod;
        x+=mod;
    }else{
        if(x>=mod) x%=mod;
    }
    return x;
}
ll mul(ll a,ll b){
	ll ans = (a*b)%mod;
	if(ans<0) ans+=mod;
	return ans;
}
typedef tree<int,null_type,less<ll>,rb_tree_tag,tree_order_statistics_node_update> ordered_set;
typedef tree<int,null_type,less_equal<ll>,rb_tree_tag,tree_order_statistics_node_update> ordered_multiset;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
ll rnd(ll l,ll r){
    return uniform_int_distribution<ll>(l,r)(rng);
}
*/
#define maxn 100005
#define maxx 3000005
ll n,m;
struct tupl{
    ll t,l,r,c;
} a[maxn];
ll tsz = 1,tsz2 = 1;
ll ls[maxx],rs[maxx];
ll ls2[maxx],rs2[maxx];
set<pll> t[maxx],t2[maxn];
void upd(ll v,ll tl,ll tr,ll i,pll p,bool f){
    if(f) t[v].insert(p);
    else t[v].erase(p);
    if(tl==tr) return;
    ll mid = (tl+tr)/2;
    if(i<=mid){
        if(!ls[v]) ls[v] = ++tsz;
        upd(ls[v],tl,mid,i,p,f);
    }else{
        if(!rs[v]) rs[v] = ++tsz;
        upd(rs[v],mid+1,tr,i,p,f);
    }
}
void upd2(ll v,ll tl,ll tr,ll i,pll p,bool f){
    if(f) t2[v].insert(p);
    else t2[v].erase(p);
    if(tl==tr) return;
    ll mid = (tl+tr)/2;
    if(i<=mid){
        if(!ls2[v]) ls2[v] = ++tsz2;
        upd2(ls2[v],tl,mid,i,p,f);
    }else{
        if(!rs2[v]) rs2[v] = ++tsz2;
        upd2(rs2[v],mid+1,tr,i,p,f);
    }
}
set<ll> s;
void get(ll v,ll tl,ll tr,ll l,ll r,ll x){
    if(l>r) return;
    if(tl==l&&tr==r){
        auto it = t[v].begin();
        while(it!=t[v].end()&&(*it).fi<=x){
            s.insert((*it).sc);
            it++;
        }
        return;
    }
    ll mid = (tl+tr)/2;
    if(ls[v]) get(ls[v],tl,mid,l,min(mid,r),x);
    if(rs[v]) get(rs[v],mid+1,tr,max(mid+1,l),r,x);
}
void get2(ll v,ll tl,ll tr,ll l,ll r,ll x){
    if(l>r) return;
    if(tl==l&&tr==r){
        auto it = t2[v].begin();
        while(it!=t2[v].end()&&(*it).fi<=x){
            s.insert((*it).sc);
            it++;
        }
        return;
    }
    ll mid = (tl+tr)/2;
    if(ls2[v]) get2(ls2[v],tl,mid,l,min(mid,r),x);
    if(rs2[v]) get2(rs2[v],mid+1,tr,max(mid+1,l),r,x);
}
ll d[maxn];
void tc(){
    cin >> m >> n;
    for(ll i = 1;i<=n;i++) cin >> a[i].t >> a[i].l >> a[i].r >> a[i].c;
    for(ll i = 1;i<=n;i++) a[i].l--;
    ll mxt = 0;
    for(ll i = 1;i<=n;i++) mxt = max(mxt,a[i].t);
    for(ll i = 1;i<=n;i++){
        upd(1,1,mxt,a[i].t,{a[i].l+a[i].t,i},1);
        upd2(1,1,mxt,a[i].t,{a[i].l-a[i].t,i},1);
    }
    for(ll i = 1;i<=n;i++) d[i] = llinf;
    priority_queue<pll> pq;
    for(ll i = 1;i<=n;i++){
        if(a[i].l==0){
            d[i] = a[i].c;
            pq.push({-d[i],i});
            upd(1,1,mxt,a[i].t,{a[i].l+a[i].t,i},0);
            upd2(1,1,mxt,a[i].t,{a[i].l-a[i].t,i},0);
        }
    }
    while(sz(pq)){
        pll p = pq.top();
        ll i = p.sc;
        pq.pop();
        if(-p.fi!=d[i]) continue;
        s.clear();
        get(1,1,mxt,a[i].t,mxt,a[i].r+a[i].t);
        get2(1,1,mxt,1,a[i].t,a[i].r-a[i].t);
        for(ll j : s){
            if(j==i) continue;
            d[j] = d[i] + a[j].c;
            upd(1,1,mxt,a[j].t,{a[j].l+a[j].t,j},0);
            upd2(1,1,mxt,a[j].t,{a[j].l-a[j].t,j},0);
            pq.push({-d[j],j});
        }
    }
    ll ans = llinf;
    for(ll i = 1;i<=n;i++) if(a[i].r==m) ans = min(ans,d[i]);
    if(ans==llinf) ans = -1;
    cout<<ans<<endl;
}
int main(){
	daj_mi_malo_vremena
    int t; t = 1;
    while(t--){
        tc();
    }
	return 0;
}

• Subtask 1: 4 / 4
• Subtask 2: 5 / 5
• Subtask 3: 30 / 30
• Subtask 4: 0 / 61