P2048 超级钢琴（RMQ）

目录

• Description
• State
• Input
• Output
• Solution
• Code

Description

从 n 个数中，挑选出 k 个长度为 [L, R] 的子段，使得他们的子段之和最大，输出总的最大值

State

\(1<=n<=5*10^{5}\)

\(1<=k<=5*10^5\)

\(1<=L<=R<=n\)

\(-1000<=a[i]<=1000\)

Input

4 3 2 3
3
2
-6
8

Output

11

Solution

想要枚举所有满足长度的子段显然是不合理的，但是对于每一个 \(i\) 都有 \(x∈[i + L - 1, i +R - 1]\) ，使得 \(\sum_{i}^xa[i]\) 最大，下一次的时候将区间分为 \(x∈[L,x-1]\) 和 \(x∈[x+1,R]\) 这两个区间就可以了

Code

const int N = 5e5 + 5;

    ll n, m, _;
    int i, j, k;
    int a[N];
    int sum[N];
    struct Node
    {
        int l, r;
        int maxx;
        int id;
        #define lson id << 1
        #define rson id << 1 | 1
        Node(){}
        Node(int maxx):maxx(maxx){}
        friend Node max(Node a, Node b){
            Node ans(-2e9);
            if(a.maxx > b.maxx){
                ans = a;
            }   
            else{
                ans = b;
            }
            return ans;
        }
    }t[N << 2];

void push_up(int id)
{
    int L = t[id].l, R = t[id].r;
    t[id] = max(t[lson], t[rson]);
    t[id].l = L, t[id].r = R;
}

void build(int l, int r, int id)
{
    t[id].l = l, t[id].r = r;
    t[id].maxx = -2e9;
    if(l == r){
        t[id].id = l;
        t[id].maxx = sum[l];
        return ;
    }
    else{
        int mid = l + r >> 1;
        build(l, mid, lson);
        build(mid + 1, r, rson);
        push_up(id);
    }
}

Node query(int l, int r, int id)
{
    int L = t[id].l, R = t[id].r;
    if(L >= l && r >= R){
        return t[id];
    }
    else{
        Node ans(-2e9);
        int mid = L + R >> 1;
        if(mid >= l) ans = max(ans, query(l, r, lson));
        if(r >= mid + 1) ans = max(ans, query(l, r, rson));
        return ans;
    }
}

struct Que
{
    int aid, bid, val;
    int l, r;
    bool operator<(Que o) const{
        return val < o.val;
    }
    Que(){}
    Que(int id, int l, int r) : aid(id), l(l), r(r){
        Node ans = query(l, r, 1);
        bid = ans.id;
        val = ans.maxx - sum[aid - 1];
    }
};

signed main()
{
    //IOS;
    int l, r;
    while(~ sdd(n, k)){
        sdd(l, r);
        rep(i, 1, n){
            sd(a[i]);
            sum[i] = sum[i - 1] + a[i];
        }
        build(1, n, 1);
        priority_queue<Que> q;
        rep(i, 1, n - l + 1){
            q.push(Que(i, i + l - 1, min(i + r - 1ll, n)));
        }
        ll ans = 0;
        rep(i, 1, k){
            Que top = q.top();
            q.pop();
            int aid = top.aid, bid = top.bid, L = top.l, R = top.r;
            ans += top.val; 
            if(bid != L){
                q.push(Que(aid, L, bid - 1));
            }
            if(bid != R){
                q.push(Que(aid, bid + 1, R));
            }
        }
        pll(ans);
    }
    return 0;
}