BOJ 14727 : 퍼즐 자르기

• 문제 링크 : boj.kr/14727
• 난이도 : P5
• 태그 : 세그먼트 트리, 분할 정복

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

        ll ans = w*h;
        ans = max(ans, find(left, q.first-1));
        ans = max(ans, find(q.first+1, right));
        return ans;
    };
    cout << find(1, n) << "\n";

    return 0;
}

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풀이

#6549와 같은 문제이다.