Submission #590458

#	Time	Username	Problem	Language	Result	Execution time	Memory
590458		Lobo	The short shank; Redemption (BOI21_prison)	C++17	80 / 100	2097 ms	114472 KiB

prison

#include<bits/stdc++.h>
using namespace std;
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
 
#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")
const int inf1 = (int) 1e9 + 10;
// #define int long long
#define dbl long double
#define endl '\n'
#define sc second
#define fr first
#define mp make_pair
#define pb push_back
#define all(x) x.begin(), x.end()
 
const int maxn = 2e6+10;
 
int n, d, T, t[maxn], r[maxn], trT[4*maxn], lzqtd[4*maxn];
pair<int,int> trqtd[4*maxn], trr[4*maxn];
 
void buildT(int no, int l, int r) {
    trT[no] = inf1;
    if(l != r) {
        int f1=((no)<<1);
        int f2=((no)<<1)+1;
        int mid = (l+r)>>1;
        buildT(f1,l,mid);
        buildT(f2,mid+1,r);
    }
}
 
void attT(int no, int l, int r, int pos, int val) {
    if(l > pos || r < pos) return;
    if(l == r) {
        trT[no] = val;
        return;
    }
    int f1=((no)<<1);
    int f2=((no)<<1)+1;
    int mid = (l+r)>>1;
    attT(f1,l,mid,pos,val);
    attT(f2,mid+1,r,pos,val);
    trT[no] = min(trT[f1],trT[f2]);
}
 
int findT(int no, int l, int r, int val) {
    if(l == r) return l;
    int f1 = ((no)<<1);
    int f2 = ((no)<<1)+1;
    int mid = (l+r)>>1;
    if(trT[f2] <= val) return findT(f2,mid+1,r,val);
    return findT(f1,l,mid,val);
}
 
void buildqtd(int no, int l, int r) {
    if(l == r) {
        trqtd[no] = mp(0,l);
        return;
    }
    int f1=((no)<<1);
    int f2=((no)<<1)+1;
    int mid = (l+r)>>1;
    buildqtd(f1,l,mid);
    buildqtd(f2,mid+1,r);
    trqtd[no] = max(trqtd[f1],trqtd[f2]);
}
 
void flushqtd(int no, int l, int r) {
    trqtd[no].fr+= lzqtd[no];
    if(l != r) {
        lzqtd[((no)<<1)]+=lzqtd[no];
        lzqtd[((no)<<1)+1]+=lzqtd[no];
    }
    lzqtd[no] = 0;
}
 
//quero pegar o cara com maior qtd em cada momento
void attqtd(int no, int l, int r, int L, int R, int val) {
    if(lzqtd[no] != 0) flushqtd(no,l,r);
    if(l > R || r < L) return;
    if(l >= L && r <= R) {
        lzqtd[no] = val;
        flushqtd(no,l,r);
        return;
    }
    int f1=((no)<<1);
    int f2=((no)<<1)+1;
    int mid = (l+r)>>1;
    attqtd(f1,l,mid,L,R,val);
    attqtd(f2,mid+1,r,L,R,val);
    trqtd[no] = max(trqtd[f1],trqtd[f2]);
}
 
void attr(int no, int l, int r, int pos, int val) {
    if(l > pos || r < pos) return;
    if(l == r) {
        trr[no] = mp(val,pos);
        return;
    }
    int f1=((no)<<1);
    int f2=((no)<<1)+1;
    int mid = (l+r)>>1;
    attr(f1,l,mid,pos,val);
    attr(f2,mid+1,r,pos,val);
    trr[no] = min(trr[f1],trr[f2]);
}
 
pair<int,int> qrrr(int no, int l, int r, int L, int R) {
    if(l > R || r < L || L > R) return mp(n+1,0);
    if(l >= L && r <= R) return trr[no];
    int mid = (l+r)>>1;
    return min(qrrr(((no)<<1),l,mid,L,R),qrrr(((no)<<1)+1,mid+1,r,L,R));
}
 
int32_t main() {
    ios::sync_with_stdio(false); cin.tie(0);
 
    cin >> n >> d >> T;
    d = min(d,n-1);
    int ans = 0;
    for(int i = 1; i <= n; i++) cin >> t[i];
    buildT(1,1,n);
    buildqtd(1,1,n);
    for(int i = 1; i <= n; i++) {
        attT(1,1,n,i,t[i]-i);
        if(trT[1] > T-i) {
            r[i] = n+1;
        }
        else {
            r[i] = findT(1,1,n,T-i);
            ans++;
        }
        attr(1,1,n,i,r[i]);
        //i aumenta o qtd de r[i] até i-1
        if(r[i] != i && r[i] != n+1) {
            attqtd(1,1,n,r[i],i-1,1);
        }
 
        // cout << i << " " << r[i] << endl;
    }
 
    while(d--) {
        flushqtd(1,1,n);
        int id = trqtd[1].sc;
        if(trqtd[1].fr == 0) break;
        //tira todos os j que tem r[j] <= id
        //pego os com menor r[j] de id+1 até n ate que seja maior que id
        while(true) {
            auto qr = qrrr(1,1,n,id+1,n);
            if(qr.fr > id) break;
            int j = qr.sc;
            attqtd(1,1,n,qr.fr,j-1,-1);
            attr(1,1,n,j,n+1);
            ans--;
        }
    }
    cout << ans << endl;
 
}

• Subtask 1: 15 / 15
• Subtask 2: 10 / 10
• Subtask 3: 20 / 20
• Subtask 4: 10 / 10
• Subtask 5: 25 / 25
• Subtask 6: 0 / 20