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Submission #1032632

# Submission time Handle Problem Language Result Execution time Memory
1032632 2024-07-24T04:43:08 Z caterpillow Exam (eJOI20_exam) C++17
65 / 100
 88 ms 98888 KB 
#include <bits/stdc++.h>

using namespace std;

using db = long double;
using ll = long long;
using pl = pair<ll, ll>;
using pi = pair<int, int>;
#define vt vector
#define f first
#define s second
#define pb push_back
#define all(x) x.begin(), x.end() 
#define size(x) ((int) (x).size())
#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define ROF(i, a, b) for (int i = (b) - 1; i >= (a); i--)
#define F0R(i, b) FOR (i, 0, b)
#define endl '\n'
const ll INF = 1e18;
const int inf = 1e9;

template<typename... Args> // tuples
ostream& operator<<(ostream& os, tuple<Args...> t) { 
    apply([&](Args... args) { string dlm = "{"; ((os << dlm << args, dlm = ", "), ...); }, t);
    return os << "}";
}

template<typename T, typename V> // pairs
ostream& operator<<(ostream& os, pair<T, V> p) { return os << "{" << p.f << ", " << p.s << "}"; }

template<class T, class = decltype(begin(declval<T>()))> // iterables
typename enable_if<!is_same<T, string>::value, ostream&>::type operator<<(ostream& os, const T& v) {
    string dlm = "{";
    for (auto& i : v) os << dlm << i, dlm = ", ";
    return os << "}";  
}

template <typename T, typename... V>
void printer(string pfx, const char *names, T&& head, V&& ...tail) {
    int i = 0;
    while (names[i] && names[i] != ',') i++;
    constexpr bool is_str = is_same_v<decay_t<T>, const char*>;
    if (is_str) cerr << " " << head;
    else cerr << pfx, cerr.write(names, i) << " = " << head; 
    if constexpr (sizeof...(tail)) printer(is_str ? "" : ",", names + i + 1, tail...);
    else cerr << endl;
}

#ifdef LOCAL
#define dbg(...) printer(to_string(__LINE__) + ": ", #__VA_ARGS__, __VA_ARGS__)
#else
#define dbg(x...)
#define cerr if (0) std::cerr
#endif

tuple<int, vt<int>, vt<int>> read() {
    int n; cin >> n;
    vt<int> a(n), b(n), c;
    F0R (i, n) cin >> a[i], c.pb(a[i]);
    F0R (i, n) cin >> b[i], c.pb(b[i]);
    sort(all(c));
    c.erase(unique(all(c)), c.end());
    auto ord = [&] (int x) { return lower_bound(all(c), x) - c.begin(); };
    for (auto& x : a) x = ord(x);
    for (auto& x : b) x = ord(x);
    return {n, a, b};
}

/*

we need to figure out what final configurations of A are possible
a person's level can only increase

rules: 
    1. some dude's work will always be a continuous range
    2. some dude's work can only spread up to the next larger guy on either side
    3. left to right ordering must be the same

as long as final state assignments follow these, its possible to construct

subtask 2: B is equal
    bruh

subtask 4: elements of A are distinct
    if kid i was to pass, then we know who gave him their work
    let dp[i][j] be the max passes, only allowing the first i kids to spread their work, considering only the first j kids in the line

everything else: n <= 5000
    let dp[i][j] be the max value of assigning some work to the first j kids, using only the first i kids' work
    for each kid, they are allowed to "spread" their range up to the next larger element on either side
    for each prefix, starting from the first kid he can claim, find the new max value of the prefix, additionally allowing the kid to claim some suffix of it

*/

int n;
vt<int> a, b;

struct SegTree {
    int n;
    vt<int> seg;
    void init(int _n) {
        for (n = 1; n < _n; n *= 2);
        seg.resize(2 * n);
    }
    void chmax(int l, int r, int v) {
        dbg(l, r, v);
        for (l += n, r += n + 1; l < r; l /= 2, r /= 2) {
            if (l & 1) seg[l] = max(seg[l], v), l++;
            if (r & 1) r--, seg[r] = max(seg[r], v);
        }
    }
    int operator[](int i) {
        int res = seg[i += n];
        while (i /= 2) res = max(res, seg[i]);
        return res;
    }
};

void solve3() {
    vt<int> rnum(2 * n, -1);
    F0R (i, n) rnum[a[i]] = i;

    vt<pi> rng(n);
    vt<int> stk;
    F0R (i, n + 1) {
        while (size(stk) && (i == n || a[stk.back()] < a[i])) {
            rng[stk.back()].s = i;
            stk.pop_back();
        }
        stk.pb(i);
    }
    stk.clear();
    ROF (i, -1, n) {
        while (size(stk) && (i == -1 || a[stk.back()] < a[i])) {
            rng[stk.back()].f = i;
            stk.pop_back();
        }
        stk.pb(i);
    }

    vt<vt<int>> nums(2 * n);
    F0R (i, n) {
        int j = rnum[b[i]];
        if (j == -1) continue;
        if (rng[j].f <= i && i <= rng[j].s) {
            nums[b[i]].pb(i);
        }
    }

    SegTree dp;
    dp.init(n + 1);
    dbg(a);
    dbg(b);
    dbg(rnum);
    F0R (i, n) {
        if (rnum[b[i]] == -1) continue; 
        int m = size(nums[a[i]]);
        dbg(m, nums[a[i]]);
        FOR (j, 0, m) {
            dp.chmax(nums[a[i]][j] + 1, n, dp[nums[a[i]][j]] + 1);
        }
    }
    cout << dp[n] << endl;
}

void solve2() {
    int r = -1;
    int ans = 0;
    F0R (i, n) {
        if (a[i] == b[0]) {
            int j = i;
            while (j > r && a[j] <= a[i]) {
                j--;
                ans++;
            }
            j = i + 1;
            r = i;
            while (a[j] < n && a[j] <= a[i]) {
                ans++;
                j++;
                r++;
            }
        }
    }
    cout << ans << endl;
}

void solve1() {
    vt<vt<int>> dp(n + 1, vt<int>(n + 1)); // 1-indexed
    vt<pi> rng(n); // reachable range
    F0R (i, n) {
        int j = i;
        while (j < n && a[j] <= a[i]) j++;
        rng[i].s = j - 1;
        j = i;
        while (j >= 0 && a[j] <= a[i]) j--;
        rng[i].f = j + 1;
    }
    dbg(rng);

    // dp
    F0R (i, n) {
        dbg(dp[i]);
        F0R (j, n + 1) dp[i + 1][j] = dp[i][j];
        auto [l, r] = rng[i];
        int best = dp[i][l];
        FOR (j, l, r + 1) {
            best = max(best + (b[j] == a[i]), dp[i][j + 1]);
            dp[i + 1][j + 1] = best;
        }
    }
    cout << dp[n][n] << endl;
}

main() {
    cin.tie(0)->sync_with_stdio(0);
    
    tie(n, a, b) = read();
    if (n <= 5000) {
        solve1();
    } else {
        int v = b[0];
        bool is_same = true;
        FOR (i, 1, n) is_same &= b[i] == v;
        if (is_same) solve2();
        else solve3();
    }
}

Compilation message

exam.cpp:215:1: warning: ISO C++ forbids declaration of 'main' with no type [-Wreturn-type]
  215 | main() {
      | ^~~~

Subtask #1
14.0 / 14.0

# Verdict Execution time Memory Grader output
1 Correct 0 ms 344 KB Output is correct
2 Correct 0 ms 348 KB Output is correct
3 Correct 0 ms 348 KB Output is correct
4 Correct 0 ms 348 KB Output is correct
5 Correct 0 ms 348 KB Output is correct
6 Correct 1 ms 344 KB Output is correct

Subtask #2
0 / 12.0

# Verdict Execution time Memory Grader output
1 Correct 3 ms 4444 KB Output is correct
2 Incorrect 11 ms 1372 KB Output isn't correct
3 Halted 0 ms 0 KB -

Subtask #3
13.0 / 13.0

# Verdict Execution time Memory Grader output
1 Correct 0 ms 344 KB Output is correct
2 Correct 1 ms 1372 KB Output is correct
3 Correct 12 ms 16092 KB Output is correct
4 Correct 66 ms 90736 KB Output is correct
5 Correct 70 ms 98640 KB Output is correct
6 Correct 70 ms 98644 KB Output is correct

Subtask #4
0 / 23.0

# Verdict Execution time Memory Grader output
1 Incorrect 2 ms 860 KB Output isn't correct
2 Halted 0 ms 0 KB -

Subtask #5
16.0 / 16.0

# Verdict Execution time Memory Grader output
1 Correct 0 ms 344 KB Output is correct
2 Correct 0 ms 348 KB Output is correct
3 Correct 0 ms 348 KB Output is correct
4 Correct 0 ms 348 KB Output is correct
5 Correct 0 ms 348 KB Output is correct
6 Correct 1 ms 344 KB Output is correct
7 Correct 0 ms 348 KB Output is correct
8 Correct 0 ms 348 KB Output is correct
9 Correct 0 ms 456 KB Output is correct
10 Correct 1 ms 600 KB Output is correct
11 Correct 0 ms 604 KB Output is correct
12 Correct 0 ms 468 KB Output is correct
13 Correct 1 ms 604 KB Output is correct
14 Correct 0 ms 600 KB Output is correct

Subtask #6
22.0 / 22.0

# Verdict Execution time Memory Grader output
1 Correct 0 ms 344 KB Output is correct
2 Correct 0 ms 348 KB Output is correct
3 Correct 0 ms 348 KB Output is correct
4 Correct 0 ms 348 KB Output is correct
5 Correct 0 ms 348 KB Output is correct
6 Correct 1 ms 344 KB Output is correct
7 Correct 0 ms 344 KB Output is correct
8 Correct 1 ms 1372 KB Output is correct
9 Correct 12 ms 16092 KB Output is correct
10 Correct 66 ms 90736 KB Output is correct
11 Correct 70 ms 98640 KB Output is correct
12 Correct 70 ms 98644 KB Output is correct
13 Correct 0 ms 348 KB Output is correct
14 Correct 0 ms 348 KB Output is correct
15 Correct 0 ms 456 KB Output is correct
16 Correct 1 ms 600 KB Output is correct
17 Correct 0 ms 604 KB Output is correct
18 Correct 0 ms 468 KB Output is correct
19 Correct 1 ms 604 KB Output is correct
20 Correct 0 ms 600 KB Output is correct
21 Correct 0 ms 600 KB Output is correct
22 Correct 2 ms 4444 KB Output is correct
23 Correct 88 ms 98888 KB Output is correct
24 Correct 68 ms 98644 KB Output is correct
25 Correct 55 ms 98492 KB Output is correct
26 Correct 53 ms 98392 KB Output is correct
27 Correct 51 ms 98384 KB Output is correct
28 Correct 53 ms 98644 KB Output is correct
29 Correct 68 ms 98640 KB Output is correct
30 Correct 69 ms 98640 KB Output is correct
31 Correct 82 ms 98644 KB Output is correct
32 Correct 62 ms 98640 KB Output is correct