Submission #405812

#	Time	Username	Problem	Language	Result	Execution time	Memory
405812		534351	The short shank; Redemption (BOI21_prison)	C++17	80 / 100	2028 ms	163744 KiB

prison

#include <bits/stdc++.h>

using namespace std;

template<class T, class U>
void ckmin(T &a, U b)
{
    if (a > b) a = b;
}

template<class T, class U>
void ckmax(T &a, U b)
{
    if (a < b) a = b;
}

#define MP make_pair
#define PB push_back
#define LB lower_bound
#define UB upper_bound
#define fi first
#define se second
#define FOR(i, a, b) for (auto i = (a); i < (b); i++)
#define FORD(i, a, b) for (auto i = (a) - 1; i >= (b); i--)
#define SZ(x) ((int) (x).size())
#define ALL(x) (x).begin(), (x).end()

const int MAXN = 2100013;
const long long LLINF = 3e18;

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<pii> vpi;
typedef vector<pll> vpl;

int N, D, T;
int arr[MAXN], lt[MAXN];
int fen[MAXN];
ll dp[MAXN];
int cost[MAXN], taken[MAXN];
int ans;

void update(int idx, int v)
{
    for (int e = idx + 1; e <= N; e += e & (-e))
    {
        ckmax(fen[e], v);
    }
}
int query(int idx)
{
    int res = -1;
    for (int e = idx + 1; e; e -= e & (-e))
    {
        ckmax(res, fen[e]);
    }
    return res;
}
//maintain a segment tree of (dp, best dp)

pll seg[2 * MAXN]; //store (min, argmin)
ll lazy[2 * MAXN];

void build(int w, int L, int R)
{
    if (L == R)
    {
        seg[w] = {LLINF, L};
        lazy[w] = 0;
        return;
    }
    int mid = (L + R) >> 1;
    build(w << 1, L, mid);
    build(w << 1 | 1, mid + 1, R);
    seg[w] = min(seg[w << 1], seg[w << 1 | 1]);
    lazy[w] = 0;
}
void push(int w, int L, int R)
{
    if (lazy[w] == 0) return;
    seg[w].fi += lazy[w];
    if (L != R)
    {
        lazy[w << 1] += lazy[w];
        lazy[w << 1 | 1] += lazy[w];
    }
    lazy[w] = 0;
    return;
}
void update(int w, int L, int R, int a, int b, ll v)
{
    push(w, L, R);
    if (b < L || R < a) return;
    if (a <= L && R <= b)
    {
        lazy[w] += v;
        push(w, L, R);
        return;
    }
    int mid = (L + R) >> 1;
    update(w << 1, L, mid, a, b, v);
    update(w << 1 | 1, mid + 1, R, a, b, v);
    seg[w] = min(seg[w << 1], seg[w << 1 | 1]);
}

void check(int c)
{
    // cerr << "CHECKING " << c << endl;
    //assign a penalty of c for starting a new range at c.
    //minimum # of guys rebelling.
    build(1, 0, N);
    dp[0] = 0; taken[0] = 0;
    update(1, 0, N, 0, 0, -LLINF);
    // fill(cost, cost + N + 1, 0);
    FOR(i, 0, N)
    {
        int v = lt[i];
        update(1, 0, N, 0, v, 1);
        // FOR(j, 0, v + 1)
        // {
        //     cost[j]++;
        // }
        // FOR(j, 0, i + 1)
        // {
        //     if (dp[i + 1] > dp[j] + cost[j] + c)
        //     {
        //         dp[i + 1] = dp[j] + cost[j] + c;
        //         taken[i + 1] = taken[j] + 1;
        //     }
        // }
        dp[i + 1] = seg[1].fi + c;
        taken[i + 1] = taken[seg[1].se] + 1;
        update(1, 0, N, i + 1, i + 1, -LLINF + dp[i + 1]);
        // cerr << dp[i + 1] << ' ' << taken[i + 1] << endl;
    }
    // cerr << "DP " << dp[N] << " TAKEN " << c << endl;
    return;
}

int32_t main()
{
    ios_base::sync_with_stdio(false); cin.tie(0);
    cout << fixed << setprecision(12);
    cerr << fixed << setprecision(4);
    cin >> N >> D >> T; D++;
    FOR(i, 0, N)
    {
        cin >> arr[i];
    }
    fill(fen, fen + N + 1, -1);
    FOR(i, 0, N)
    {
        int c = arr[i];
        if (c <= T)
        {
            //i...i+(T-c) are getting infected.
            update(max(0, N - 1 - (i + (T - c))), i);
        }
        lt[i] = query(N - 1 - i);
        // cerr << lt[i] << ' ';
    }
    if (D == 2)
    {
        FOR(i, 0, N)
        {
            int v = lt[i];
            // cerr << v << endl;
            //it will rebel if:
            //v != -1
            //the block is in 0...v-1 or i...N-1
            if (v != -1)
            {
                cost[0]++;
                cost[v]--;
                cost[i]++;
                cost[N]--;
            }
        }
        FOR(i, 0, N)
        {
            cost[i + 1] += cost[i];
        }
        ans = N;
        FOR(i, 0, N - 1)
        {
            ckmin(ans, cost[i]);
        }
        cout << ans << '\n';
        return 0;
    }
    int lo = 0, hi = N + 1;
    while(hi > lo)
    {
        int mid = (hi + lo + 1) >> 1;
        check(mid);
        if (taken[N] >= D) lo = mid;
        else hi = mid - 1;
    }
    // cerr << "LO = " << lo << endl;
    check(lo);
    // cerr << "TAKED " << taken[N] << endl;
    ll cost1 = dp[N] - lo * taken[N]; int cnt1 = taken[N];
    assert(cnt1 >= D);
    if (cnt1 == D)
    {
        ans = cost1;
    }
    else
    {
        hi++;
        check(hi);
        ll cost2 = dp[N] - hi * taken[N]; int cnt2 = taken[N];
        assert(cnt2 < D);
        // cerr << cost1 << ' ' << cnt1 << ' ' << cost2 << ' ' << cnt2 << endl;
        ans = cost2 + (cost1 - cost2) * (D - cnt2) / (cnt1 - cnt2);
    }
    cout << ans << '\n';
    //you are going to have some new blocks starting with -1....x. and then -1...x and then -1...x
    // cerr << endl;
    return 0;
}

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