제출 #1005404

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1005404		GrindMachine	The short shank; Redemption (BOI21_prison)	C++17	100 / 100	1709 ms	377388 KiB

prison

#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx2")

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
 
using namespace std;
using namespace __gnu_pbds;
 
template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long int ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
 
#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL)
#define pb push_back
#define endl '\n'
#define sz(a) (int)a.size()
#define setbits(x) __builtin_popcountll(x)
#define ff first
#define ss second
#define conts continue
#define ceil2(x,y) ((x+y-1)/(y))
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define yes cout << "Yes" << endl
#define no cout << "No" << endl
 
#define rep(i,n) for(int i = 0; i < n; ++i)
#define rep1(i,n) for(int i = 1; i <= n; ++i)
#define rev(i,s,e) for(int i = s; i >= e; --i)
#define trav(i,a) for(auto &i : a)
 
template<typename T>
void amin(T &a, T b) {
    a = min(a,b);
}
 
template<typename T>
void amax(T &a, T b) {
    a = max(a,b);
}
 
#ifdef LOCAL
#include "debug.h"
#else
#define debug(x) 42
#endif
 
/*
 
refs:
edi
 
*/
 
const int MOD = 1e9 + 7;
const int N = 1e5 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;
 
template<typename T>
struct sparse_table {
    /*============================*/
 
    T merge(T a, T b) {
        return min(a,b);
    }
 
    /*============================*/
 
    vector<vector<T>> table;
    vector<int> bin_log;
    int LOG = 0;
 
    sparse_table() {
 
    }
 
    sparse_table(vector<T> &a, int n) {
        while ((1 << LOG) <= n) LOG++;
 
        table = vector<vector<T>>(n, vector<T>(LOG));
        bin_log = vector<int>(n + 1);
 
        rep(i, n) table[i][0] = a[i];
 
        rep1(j, LOG - 1) {
            rep(i, n) {
                int jump = 1 << (j - 1);
                if (i + jump >= n) {
                    break;
                }
 
                table[i][j] = merge(table[i][j - 1], table[i + jump][j - 1]);
            }
        }
 
        bin_log[1] = 0;
        for (int i = 2; i <= n; ++i) {
            bin_log[i] = bin_log[i / 2] + 1;
        }
    }
 
    T query(int l, int r) {
        int len = r - l + 1;
        int k = bin_log[len];
 
        T val1 = table[l][k];
        T val2 = table[r - (1 << k) + 1][k];
 
        return merge(val1, val2);
    }
};

template<typename T>
struct lazysegtree {
    /*=======================================================*/

    struct data {
        int mx,ind;
    };

    struct lazy {
        int a;
    };

    data d_neutral = {-inf1,-1};
    lazy l_neutral = {0};

    void merge(data &curr, data &left, data &right) {
        if(left.mx >= right.mx) curr = left;
        else curr = right;
    }

    void create(int x, int lx, int rx, T v) {
        tr[x] = {v,lx};
    }

    void modify(int x, int lx, int rx, T v) {
        lz[x].a = v;
    }

    void propagate(int x, int lx, int rx) {
        int v = lz[x].a;
        if(!v) return;

        tr[x].mx += v;

        if(rx-lx > 1){
            lz[x<<1].a += v;
            lz[x<<1|1].a += v;
        }

        lz[x] = l_neutral;
    }

    /*=======================================================*/

    int siz = 1;
    vector<data> tr;
    vector<lazy> lz;

    lazysegtree() {

    }

    lazysegtree(int n) {
        while (siz < n) siz *= 2;
        tr.assign(2 * siz, d_neutral);
        lz.assign(2 * siz, l_neutral);
    }

    void build(vector<T> &a, int n, int x, int lx, int rx) {
        if (rx - lx == 1) {
            if (lx < n) {
                create(x, lx, rx, a[lx]);
            }

            return;
        }

        int mid = (lx + rx) >> 1;

        build(a, n, x<<1, lx, mid);
        build(a, n, x<<1|1, mid, rx);

        merge(tr[x], tr[x<<1], tr[x<<1|1]);
    }

    void build(vector<T> &a, int n) {
        build(a, n, 1, 0, siz);
    }

    void rupd(int l, int r, T v, int x, int lx, int rx) {
        propagate(x, lx, rx);

        if (lx >= r or rx <= l) return;
        if (lx >= l and rx <= r) {
            modify(x, lx, rx, v);
            propagate(x, lx, rx);
            return;
        }

        int mid = (lx + rx) >> 1;

        rupd(l, r, v, x<<1, lx, mid);
        rupd(l, r, v, x<<1|1, mid, rx);

        merge(tr[x], tr[x<<1], tr[x<<1|1]);
    }

    void rupd(int l, int r, T v) {
        rupd(l, r + 1, v, 1, 0, siz);
    }

    data query(int l, int r, int x, int lx, int rx) {
        propagate(x, lx, rx);

        if (lx >= r or rx <= l) return d_neutral;
        if (lx >= l and rx <= r) return tr[x];

        int mid = (lx + rx) >> 1;

        data curr;
        data left = query(l, r, x<<1, lx, mid);
        data right = query(l, r, x<<1|1, mid, rx);

        merge(curr, left, right);
        return curr;
    }

    data query(int l, int r) {
        return query(l, r + 1, 1, 0, siz);
    }
};

template<typename T>
struct segtree {
    // https://codeforces.com/blog/entry/18051

    /*=======================================================*/

    struct data {
        int mx,ind;
    };

    data neutral = {-inf1,-1};

    data merge(data &left, data &right) {
        data curr;
        if(left.mx >= right.mx) curr = left;
        else curr = right;
        return curr;
    }

    void create(int i, T v) {
        tr[i] = {v,i-n};
    }

    void modify(int i, T v) {
        tr[i] = {v,i-n};
    }

    /*=======================================================*/

    int n;
    vector<data> tr;

    segtree() {

    }

    segtree(int siz) {
        init(siz);
    }

    void init(int siz) {
        n = siz;
        tr.assign(2 * n, neutral);
    }

    void build(vector<T> &a, int siz) {
        rep(i, siz) create(i + n, a[i]);
        rev(i, n - 1, 1) tr[i] = merge(tr[i << 1], tr[i << 1 | 1]);
    }

    void pupd(int i, T v) {
        modify(i + n, v);
        for (i = (i + n) >> 1; i; i >>= 1) tr[i] = merge(tr[i << 1], tr[i << 1 | 1]);
    }

    data query(int l, int r) {
        data resl = neutral, resr = neutral;

        for (l += n, r += n; l <= r; l >>= 1, r >>= 1) {
            if (l & 1) resl = merge(resl, tr[l++]);
            if (!(r & 1)) resr = merge(tr[r--], resr);
        }

        return merge(resl, resr);
    }
};

void solve(int test_case)
{
    int n,d,t; cin >> n >> d >> t;
    vector<int> a(n+5);
    rep1(i,n) cin >> a[i];
 
    vector<int> b(n+5);
    rep1(i,n) b[i] = a[i] - i;
 
    sparse_table<int> sparse(b,n+1);
    vector<int> lx(n+5); // lx[i] = first pos to the left of i s.t if we start making ops from this pos, a[i] <= t
    
    rep1(i,n){
        // a[j] + (i-j) = (a[j]-j) + i
        int l = 1, r = i;
        while(l <= r){
            int mid = (l+r) >> 1;
            if(sparse.query(mid,i) + i <= t){
                lx[i] = mid;
                l = mid + 1;
            }
            else{
                r = mid - 1;
            }
        }
    }
 
    vector<bool> active(n+5);
    int ignore = 0;

    rep1(i,n){
        if(a[i] > t){
            if(!lx[i]){
                ignore++;
            }
            else{
                active[i] = 1;
            }
        }
    }
    
    vector<int> diff(n+5);
    vector<pii> segs;

    rep1(i,n){
        if(!active[i]) conts;
        segs.pb({lx[i]+1,i});
    }
    sort(all(segs));
    for(auto [l,r] : segs){
        diff[l]++, diff[r+1]--;
    }
    rep1(i,n) diff[i] += diff[i-1];
    
    lazysegtree<int> st(n+5);
    st.build(diff,n+1);

    int siz = sz(segs);
    vector<int> seg_build(siz);
    rep(i,siz){
        seg_build[i] = segs[i].ss;
    }

    segtree<int> seg(siz+5);
    seg.build(seg_build,siz);

    rep1(iter,d){
        auto [mx,ind] = st.query(1,n);
        if(ind == -1) break;

        pii px = {ind+1,-1};
        int pos = upper_bound(all(segs),px)-segs.begin()-1;

        while(mx--){
            auto [mxr,ind2] = seg.query(0,pos);
            assert(mxr >= ind);
            st.rupd(segs[ind2].ff,segs[ind2].ss,-1);
            seg.pupd(ind2,-inf1);
            ignore++;
        }
    }
 
    int ans = n-ignore;
    cout << ans << endl;
}
 
int main()
{
    fastio;
 
    int t = 1;
    // cin >> t;
 
    rep1(i, t) {
        solve(i);
    }
 
    return 0;
}

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