Yukicoder No.318 学学学学学

,

http://yukicoder.me/problems/no/318

solution

区間一様代入と点取得のsegment木を書けばよい。$O((n+q)\log n)$。 segment木でなくてもできる。

implementation

#include <iostream>
#include <vector>
#include <map>
#include <cmath>
#include <functional>
#define repeat(i,n) for (int i = 0; (i) < (n); ++(i))
using namespace std;
template <class T> void setmax(T & a, T const & b) { if (a < b) a = b; }
template <class T> void setmin(T & a, T const & b) { if (b < a) a = b; }

template <typename T>
struct lazed_segment_tree { // on associative symmetric monoid
    int n;
    vector<T> a;
    function<T (T,T)> append; // associative, symmetric
    co_segment_tree() = default;
    template <typename F>
    co_segment_tree(int a_n, T a_init, F a_append) {
        n = pow(2,ceil(log2(a_n)));
        a.resize(2*n-1, a_init);
        append = a_append;
    }
    void range_update(int l, int r, T z) {
        range_update(0, 0, n, l, r, z);
    }
    void range_update(int i, int il, int ir, int l, int r, T z) {
        if (l <= il and ir <= r) {
            a[i] = append(a[i], z);
        } else if (ir <= l or r <= il) {
            // nop
        } else {
            range_update(2*i+1, il, (il+ir)/2, l, r, z);
            range_update(2*i+2, (il+ir)/2, ir, l, r, z);
        }
    }
    T point_concat(int i) {
        T z = a[i+n-1];
        for (i = (i+n)/2; i > 0; i /= 2) {
            z = append(z, a[i-1]);
        }
        return z;
    }
};

int main() {
    // input
    int n; cin >> n;
    vector<int> as(n); repeat (i,n) cin >> as[i];
    // compute
    map<int,pair<int,int> > q;
    repeat (i,n) {
        if (not q.count(as[i])) q[as[i]] = { i, i };
        setmin(q[as[i]].first,  i);
        setmax(q[as[i]].second, i);
    }
    co_segment_tree<int> t(n, 0, [](int a, int b) { return max(a, b); });
    for (auto it : q) {
        int a = it.first;
        int l, r; tie(l, r) = it.second;
        t.range_update(l, r+1, a);
    }
    // output
    repeat (i,n) {
        if (i) cout << ' ';
        cout << t.point_concat(i);
    }
    cout << endl;
    return 0;
}