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QOJ

IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#265283#7685. Barkley IIGCnoodlesbotTL 3ms20592kbC++232.1kb2023-11-25 17:40:332023-11-25 17:40:34

Judging History

你现在查看的是最新测评结果

  • [2023-11-25 17:40:34]
  • 评测
  • 测评结果:TL
  • 用时:3ms
  • 内存:20592kb
  • [2023-11-25 17:40:33]
  • 提交

answer

#include <bits/stdc++.h>

using namespace std;
using i64 = long long;

const int N = 5e5 + 10;

template<class T>
struct Fenwick {
    int n;
    vector<T> a;

    Fenwick(int n = 0) {
        init(n);
    }

    void init(int n) {
        this->n = n;
        this->a.assign(n, T());
    }

    void add(int p, T x) {
        for (; p <= n; p += p & -p) {
            a[p] += x;
        }
    }

    int sum(int p) {
        T res = 0;
        for (; p > 0; p -= p & -p) {
            res += a[p];
        }
        return res;
    }
};

vector<vector<int>> pos(N);
vector<int> pt(N), p(N), cnt(N);

void solve() {
    int n, m;
    cin >> n >> m;
    vector<int> a(n + 1);
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
        pos[a[i]].emplace_back(i);
        cnt[a[i]] = 1;
    }

    int ans = 0;
    Fenwick<int> fen(n + 1);
    for (int i = 1; i <= n; i++) {
        int mex = a[i];
        int r = i - 1, l;
        if (pos[mex][pt[mex]] == i) {
            l = 0;
        } else {
            l = pos[mex][pt[mex]];
            pt[mex]++;
        }
        cerr << l << ' ' << r << ' ' << fen.sum(r) - fen.sum(l) - mex << '\n';
        ans = max(ans, fen.sum(r) - fen.sum(l) - mex);
        if (p[mex]) {
            fen.add(p[mex], -1);
        }
        fen.add(i, 1);
        p[mex] = i;
    }
    for (int i = 1; i <= n; i++) {
        int mex = a[i];
        int r = n, l = *pos[mex].rbegin();
        cerr << l << ' ' << r << ' ' << fen.sum(r) - fen.sum(l) - mex << '\n';
        ans = max(ans, fen.sum(r) - fen.sum(l) - mex);
    }
    for (int i = 1; i < N; i++) {
        if (!cnt[i]) {
            ans = max(ans, fen.sum(n) - i);
            break;
        }
    }
    cout << ans << '\n';

    for (int i = 1; i <= n; i++) {
        pos[a[i]].clear();
        pt[a[i]] = 0;
        p[a[i]] = 0;
        cnt[a[i]] = 0;
    }
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    int t;
    cin >> t;

    while (t--) {
        solve();
    }

    return 0;
}

Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 100
Accepted
time: 3ms
memory: 20592kb

input:

2
5 4
1 2 2 3 4
5 10000
5 2 3 4 1

output:

2
3

result:

ok 2 number(s): "2 3"

Test #2:

score: -100
Time Limit Exceeded

input:

50000
10 19
12 6 1 12 11 15 4 1 13 18
10 8
8 7 6 7 6 2 2 3 4 8
10 6
3 2 6 6 5 2 3 4 5 6
10 11
6 3 7 9 2 1 2 10 10 4
10 6
6 1 2 6 1 1 3 4 2 1
10 9
8 5 3 9 1 7 5 5 1 1
10 5
1 4 3 2 5 4 5 3 5 2
10 14
3 8 12 10 4 2 3 13 7 3
10 14
5 5 12 2 8 1 13 9 8 5
10 7
5 5 6 6 1 5 3 7 3 4
10 7
5 1 4 6 1 6 4 3 7 5
10...

output:

6
5
4
4
2
4
3
7
4
4
4
5
2
3
6
6
7
5
7
6
5
5
6
2
6
8
7
2
5
5
6
2
2
3
4
5
3
3
7
3
2
5
6
1
3
5
3
3
3
8
6
6
5
7
4
4
5
4
6
6
6
3
7
3
6
3
3
7
7
6
6
7
4
3
3
4
4
6
3
4
6
6
4
5
5
9
4
5
7
5
3
5
2
2
5
6
6
8
4
3
4
5
5
5
7
7
3
2
6
5
3
5
4
4
5
6
6
5
6
7
7
4
5
7
4
7
3
7
6
6
6
5
4
2
5
4
2
3
6
5
2
6
5
5
4
3
5
6
6
6
...

result: