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IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#395086#7108. Couleurucup-team055#TL 0ms3640kbC++232.5kb2024-04-21 03:31:242024-04-21 03:31:24

Judging History

你现在查看的是最新测评结果

  • [2024-04-21 03:31:24]
  • 评测
  • 测评结果:TL
  • 用时:0ms
  • 内存:3640kb
  • [2024-04-21 03:31:24]
  • 提交

answer

#include <bits/extc++.h>
using namespace std;
using ll = long long;
using vi = vector<ll>;
const ll INF = LLONG_MAX / 4;
#define rep(i, a, b) for(ll i = a; i < (b); i++)
#define sz(a) ssize(a)
bool chmin(auto& a, auto b) { if(a <= b) return 0; a = b; return 1; }
bool chmax(auto& a, auto b) { if(a >= b) return 0; a = b; return 1; }

using namespace __gnu_pbds;
template<class T> using pbds_set = tree<T, null_type, less<>, rb_tree_tag, tree_order_statistics_node_update>;
struct T {
    pbds_set<ll> s;
    ll cost = 0;
    void push_back(ll x) {
        cost += sz(s) - s.order_of_key(x);
        s.insert(x);
    }
    void pop_back(ll x) {
        s.erase(x);
        cost -= sz(s) - s.order_of_key(x);
    }
    void pop_front(ll x) {
        s.erase(x);
        cost -= s.order_of_key(x);
    }
};
void solve() {
    ll N;
    cin >> N;
    vector<ll> A(N), P(N);
    for(ll& a : A) cin >> a;
    {
        vector<ll> idx(N);
        rep(i, 0, N) idx[i] = i;
        ranges::stable_sort(idx, {}, [&](ll i) { return A[i]; });
        rep(i, 0, N) A[idx[i]] = i;
    }
    for(ll& a : P) cin >> a;
    map<ll, ll> interval;
    multiset<ll> costs {0};
    vector<T> cc(N);
    auto add = [&](ll l, ll r, T& t) {
        if(l > r) return;
        interval.emplace(l, r);
        costs.insert(t.cost);
        swap(cc[l], t);
    };
    auto push = [&](ll l, ll r) {
        if(l > r) return;
        interval[l] = r;
        T t;
        rep(i, l, r + 1) t.push_back(A[i]);
        add(l, r, t);
    };
    auto rem = [&](ll l, ll r) {
        auto& t = cc[l];
        interval.erase(l);
        costs.erase(costs.find(t.cost));
    };
    auto erase = [&](ll i) {
        auto p = --interval.upper_bound(i);
        auto [l, r] = *p;
        T& t = cc[l];
        rem(l, r);
        if(r - i < i - l) {
            for(ll j = r; j >= i; j--) t.pop_back(A[j]);
            add(l, i - 1, t);
            push(i + 1, r);
        }
        else {
            rep(j, l, i + 1) t.pop_front(A[j]);
            add(i + 1, r, t);
            push(l, i - 1);
        }
    };
    push(0, N - 1);
    vector<ll> ans;
    ll prev = 0;
    for(ll p : P) {
        prev = *rbegin(costs);
        ans.push_back(prev);
        p ^= prev;
        p--;
        erase(p);
    }
    rep(i, 0, N) cout << ans[i] << " \n"[i + 1 == N];
}
int main() {
    cin.tie(0)->sync_with_stdio(0);
    ll T;
    cin >> T;
    while(T--) solve();
}

Details

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Test #1:

score: 100
Accepted
time: 0ms
memory: 3640kb

input:

3
5
4 3 1 1 1
5 4 5 3 1
10
9 7 1 4 7 8 5 7 4 8
21 8 15 5 9 2 4 5 10 6
15
4 8 8 1 12 1 10 14 7 14 2 9 13 10 3
37 19 23 15 7 2 10 15 2 13 4 5 8 7 10

output:

7 0 0 0 0
20 11 7 2 0 0 0 0 0 0
42 31 21 14 14 4 1 1 1 0 0 0 0 0 0

result:

ok 3 lines

Test #2:

score: -100
Time Limit Exceeded

input:

11116
10
10 5 10 3 6 4 8 5 9 8
31 27 24 11 12 3 0 2 3 1
10
8 2 7 2 8 10 1 10 9 10
6 5 2 13 2 1 0 1 3 1
10
7 10 7 6 1 3 10 6 7 9
21 18 10 1 6 5 4 8 9 10
10
2 10 4 8 8 5 7 2 6 7
20 10 9 1 15 0 4 2 9 7
10
1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9 10
10
1 2 3 4 5 6 7 8 9 10
6 3 5 2 7 10 9 1 4 8
10
1 10 1 3...

output:

21 18 16 12 10 6 4 1 1 0
12 12 10 10 4 4 4 2 1 0
20 16 9 5 3 3 3 0 0 0
22 14 8 8 5 5 2 1 1 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
19 12 7 4 4 2 2 1 0 0
20 18 8 3 1 1 0 0 0 0
45 21 21 10 3 3 3 0 0 0
17 11 8 2 1 1 1 0 0 0
13 4 1 0 0 0 0 0 0 0
29 27 22 15 9 7 4 3 1 0
26 16 9 2 1 1 1 1 1 0
0 0 0 0 0 ...

result: