제출 #1003061

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1003061		GrindMachine	The short shank; Redemption (BOI21_prison)	C++17	55 / 100	2058 ms	112976 KiB

prison

#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);
    }
};
 
void solve(int test_case)
{
    ll n,d,t; cin >> n >> d >> t;
    d++; // we need d+1 segs
    vector<ll> a(n+5);
    rep1(i,n) cin >> a[i];
 
    vector<ll> b(n+5);
    rep1(i,n) b[i] = a[i] - i;
 
    sparse_table<ll> sparse(b,n+1);
    vector<ll> 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
        ll l = 1, r = i;
        while(l <= r){
            ll 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);
    rep1(i,n){
        if(a[i] > t){
            active[i] = 1;
        }
    }

    ll ignore = 0;

    rep1(iter,d-1){
        vector<ll> diff(n+5); 
        rep1(i,n){
            if(!active[i]) conts;
            if(!lx[i]){
                ignore++;
                active[i] = 0;
                conts;
            }

            diff[lx[i]+1]++, diff[i+1]--;
        }
        rep1(i,n) diff[i] += diff[i-1];

        pll best = {-inf2,-1};
        rep1(i,n){
            pll px = {diff[i],i};
            amax(best,px);
        }

        ll pos = best.ss;
        rep1(i,n){
            if(!active[i]) conts;
            if(lx[i]+1 <= pos and pos <= i){
                ignore++;
                active[i] = 0;
            }
        }
    }

    ll 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: 0 / 25
• Subtask 6: 0 / 20