提出 #53628964

---

ソースコード 拡げる

/*
    I'm Once Again Asking For a Tree Problem.
*/

#include "bits/stdc++.h"

#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 ordered_set = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;

#define ll long long
#define ld long double
#define nl cout << '\n'
#define pri(n) cout<<fixed<<setprecision(n)
#define d5l_arr(arr, n) for(int i = 0 ; i < n ; i ++)cin >> arr[i]


void m4_htgeb_tle_kda_ya3ny() {
    cin.tie(0);
    cin.sync_with_stdio(0);
    cout.tie(0);
    cout.sync_with_stdio(0);
}


const int N = 2e5 + 10;
const ll infl = 1e18 + 10, MOD = 1e9 + 7, inf = 1e9 + 10;

int segTree[4 * N];

void update_tree(int node, int l, int r, int pos, int val) {
    if (l > pos || r <= pos)return;
    if (r - l <= 1) {
        segTree[node] = max(val, segTree[node]);
        return;
    }
    int mid = (l + r) >> 1;
    if (pos < mid)
        update_tree(2 * node + 1, l, mid, pos, val);
    else
        update_tree(2 * node + 2, mid, r, pos, val);
    segTree[node] = max(segTree[2 * node + 2], segTree[2 * node + 1]);
}

/*
 * query_tree:
 * Given ql and qr, it will return the maximum value in the range [ql,qr)
 */
int query_tree(int node, int l, int r, int ql, int qr) {
    if (l >= qr || r <= ql)return -1;
    if (l >= ql && r <= qr)return segTree[node];
    int mid = (l + r) >> 1;
    return max(query_tree(2 * node + 1, l, mid, ql, qr), query_tree(2 * node + 2, mid, r, ql, qr));
}

void solveCase() {

    int n;
    cin >> n;
    map<int, int> mp;
    int a[n];

    for (int i = 0; i < n; ++i) {
        cin >> a[i];
        mp[a[i]] = 1;
    }
    int id = 0;
    for (auto &it: mp)it.second = id++;

    vector<int> dp1(n), dp2(n);
    for (int i = 0; i < 4 * id; ++i) segTree[i] = 0;
    for (int i = 0; i < n; ++i) {
        int mx = query_tree(0, 0, id, 0, mp[a[i]]);
        dp1[i] = max(0, mx) + 1;
        update_tree(0, 0, id, mp[a[i]], dp1[i]);
    }

    for (int i = 0; i < 4 * id; ++i)segTree[i] = 0;

    for (int i = n - 1; i >= 0; i--) {
        int mx = query_tree(0, 0, id, mp[a[i]] + 1, id);
        dp2[i] = max(0, mx) + 1;
        update_tree(0, 0, id, mp[a[i]], dp2[i]);
    }
    int lis = *max_element(dp1.begin(), dp1.end());
    vector<int> ans;
    for (int i = 0; i < n; i++) {
        if (dp1[i] + dp2[i] - 1 == lis)ans.push_back(i);
    }
    cout << ans.size() << '\n';
    for (int x: ans)cout << x + 1 << " ";
}

signed main() {
    m4_htgeb_tle_kda_ya3ny();
    int test = 1;
    cin >> test;
    for (int i = 1; i <= test; ++i) {
        solveCase();
        nl;
    }
    return 0;
}

提出情報

ジャッジ結果