BOJ 1989 : 부분배열 고르기 2

• 문제 링크 : boj.kr/1989
• 난이도 : P5
• 태그 : 세그먼트 트리, 분할 정복

---

코드

#include <bits/stdc++.h>

#define all(x) (x).begin(), (x).end()

#define INF 0x7FFFFFFF

using namespace std;

using ll = long long;
using ld = long double;
using pii = pair<int,int>;
using pll = pair<ll, ll>;

template<typename T>
class segTree {
private:
    ll n;
    T id;
    T(*merge)(T, T);
    vector<T> tree;
public:
    segTree(ll n, T id, T(*merge)(T, T)) {
        this->n = n;
        this->id = id;
        this->merge = merge;
        tree.resize(n*4);
    }
    void update(ll idx, T value) {
        _update(1, 1, n, idx, value);
    }
    T query(ll l, ll r) {
        return _query(1, 1, n, l, r);
    }
private:
    void _update(int node, int s, int e, int idx, T value) {
        if(idx < s || e < idx) return;

        if(s == e) {
            tree[node] = value;
            return;
        }

        _update(node*2, s, (s+e)/2, idx, value);
        _update(node*2+1, (s+e)/2+1, e, idx, value);
        tree[node] = merge(tree[node*2], tree[node*2+1]);
    }
    T _query(int node, int s, int e, int l, int r) {
        if(l > e || r < s) return id;
        if(l <= s && e <= r) return tree[node];

        T lq = _query(node*2, s, (s+e)/2, l, r);
        T rq = _query(node*2+1, (s+e)/2+1, e, l, r);
        return merge(lq, rq);
    }
};

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

    int n;
    cin >> n;

    segTree<pll> s1(n, pll(-1, INF), [](pll a, pll b){return (a.second > b.second ? b : a);});
    segTree<ll> s2(n, 0, [](ll a, ll b){return a+b;});
    for(int i = 1; i <= n; i++) {
        int input;
        cin >> input;
        s1.update(i, {i, input});
        s2.update(i, input);
    }
    function<pair<ll, pll>(ll,ll)> find = [&](ll left, ll right)->pair<ll, pll>{
        if(left > right) return {0, {0,0}};
        pll q = s1.query(left, right);
        ll w = s2.query(left, right);
        ll h = q.second;
        ll ans = w*h;

        pair<ll, pll> ret, temp;
        ret.first = ans;
        ret.second = {left, right};

        temp = find(left, q.first-1);
        if(temp.first > ret.first) ret = temp;
        temp = find(q.first+1, right);
        if(temp.first > ret.first) ret = temp;

        return ret;
    };
    auto ans = find(1, n);
    cout << ans.first << "\n" << ans.second.first << " " << ans.second.second;

    return 0;
}

---

풀이

#2104와 비슷하게 풀 수 있다.

l과 r값만 함께 구해주면 된다.