https://codeforces.com/contest/1557/problem/D

線段樹維護dp + dp路徑記錄 + 離散化
考慮最多能保留的個數,線段樹維護相鄰兩層間轉移的1的位置的最大值.
對每層考慮,新加進來一個線段,都會產生一個非負的貢獻,每次轉移都要詢問這一層能向上轉移的每個點的最大值同時記錄轉移到上一層的id

const int maxn = 6e5 + 7;

int n, t, m;

struct node {
    int l, r, mx, id, lz;
} tr[maxn << 2];

void push_up(int p) {
    if (tr[p << 1].mx >= tr[p << 1 | 1].mx) {
        tr[p].mx = tr[p << 1].mx;
        tr[p].id = tr[p << 1].id;
    } else {
        tr[p].mx = tr[p << 1 | 1].mx;
        tr[p].id = tr[p << 1 | 1].id;
    }
}

void push_down(int p) {
    if (tr[p].lz != -1) {
        tr[p << 1].lz = tr[p << 1 | 1].lz = tr[p].lz;
        tr[p << 1].mx = tr[p << 1 | 1].mx = tr[p].lz;
        tr[p << 1].id = tr[p << 1 | 1].id = tr[p].id;
        tr[p].lz = -1;
    }
}

void build(int p, int l, int r) {
    tr[p].l = l, tr[p].r = r, tr[p].lz = -1;
    if (l == r) {
        tr[p].id = 0;
        tr[p].mx = 0;
        return;
    }
    int mi = (l + r) >> 1;
    build(p << 1, l, mi);
    build(p << 1 | 1, mi + 1, r);
    push_up(p);
}

void update(int p, int l, int r, int val, int id) {
    if (l <= tr[p].l && tr[p].r <= r) {
        tr[p].mx = val;
        tr[p].id = id;
        tr[p].lz = val;
        return;
    }
    push_down(p);
    int mi = (tr[p].l + tr[p].r) >> 1;
    if (l <= mi) update(p << 1, l, r, val, id);
    if (r > mi) update(p << 1 | 1, l, r, val, id);
    push_up(p);
}

int mx, pos;

void query(int p, int l, int r) {
    if (l <= tr[p].l && tr[p].r <= r) {
        if (tr[p].mx >= mx) {
            mx = tr[p].mx;
            pos = tr[p].id;
        }
        return;
    }
    push_down(p);
    int mi = (tr[p].l + tr[p].r) >> 1;
    if (l <= mi) query(p << 1, l, r);
    if (r > mi) query(p << 1 | 1, l, r);
}

set<int> st;
map<int, int> mp;
int cnt = 0;

int id[maxn], ls[maxn], rs[maxn];
int dp[maxn], pre[maxn];
vector<pair<int, int>> vec[maxn];

void solve() {
    cin >> n >> m;
    for (int i = 1; i <= m; i++) {
        cin >> id[i] >> ls[i] >> rs[i];
        st.insert(ls[i]), st.insert(rs[i]);
    }
    for (auto i:st) mp[i] = ++cnt;
    for (int i = 1; i <= m; i++)
        vec[id[i]].emplace_back(mp[ls[i]], mp[rs[i]]);
    build(1, 1, (int) st.size() + 1);
    for (int i = 1; i <= n; i++) {
        for (auto s:vec[i]) {
            int l = s.first, r = s.second;
            mx = 0, pos = -1;
            query(1, l, r);
            if (mx + 1 > dp[i]) {
                dp[i] = mx + 1;
                pre[i] = pos;
            }
        }
        for (auto s:vec[i]) {
            int l = s.first, r = s.second;
            update(1, l, r, dp[i], i);
        }
    }
    int mx_p = max_element(dp + 1, dp + 1 + n) - dp;
    set<int> res;
    
    for (int i = 1; i <= n; i++)
        res.insert(i);
    while (mx_p) {
        res.erase(mx_p);
        mx_p = pre[mx_p];
    }
    cout << res.size() << endl;
    for (auto i :res)
        cout << i << " ";
}