动态开点李超树（例题+板子总结）

以ABC289 G - Shopping in AtCoder store题为例

做法1：n条直线插入凸包，构建最优直线的凸包，在凸包上二分/排序后离线处理询问

做法2：动态开点李超树维护n条直线最小值，将答案再取反即为最大值，由于值域1e9，采用动态开点

板子总结

动态开点李超树，维护最小值直线，对于x的求最小值y=kx+b

rt：用于动态开点，当前点的点号

ch[][2]：用于动态开点，与01trie类似

line：直线，李超树内每个节点维护一条最优直线，与普通李超树相同

add(int &id, int L, int R, line now)：对x∈[L,R]加入直线now，形如y=kx+b，以(k,b)加入

qry(int id, int L, int R, int wz)：询问x∈[L,R]中，x=wz对应的最优直线（最小值）的y值

#include<bits/stdc++.h>
typedef long long ll;
using namespace std;
const int N=2e5+10,inf=1e9;
struct LiChaoTree{
    static const int N = 2e5 + 7, M = N * 30, inf = 1e9;
    struct line {
        ll k, b;
        line (ll kk = 0, ll bb = 0) {
            k = kk, b = bb;
        }
        ll get(ll x) {
            return x * k + b;
        }
    };
    ll inter(line aa, line bb) { // 交点的横坐标
        return (bb.b - aa.b) / (aa.k - bb.k);
    }
    int tot = 0, ch[M][2];
    line seg[M];
    void add(int &id, int L, int R, line now) {
        if(!id) return id = ++tot, seg[id] = now, void();
        int mid = (L + R) >> 1;
        ll La = seg[id].get(L), Ra = seg[id].get(R), Lb = now.get(L), Rb = now.get(R);
        if(La <= Lb && Ra <= Rb) return;
        if(La >= Lb && Ra >= Rb) return seg[id] = now, void();
        ll it = inter(seg[id], now);
        if(La <= Lb) { 
            if(it <= mid) add(ch[id][0], L, mid, seg[id]), seg[id] = now;
            else add(ch[id][1], mid + 1, R, now);
        }
        else {
            if(it <= mid) add(ch[id][0], L, mid, now);
            else add(ch[id][1], mid + 1, R, seg[id]), seg[id] = now;
        }
    }
    ll qry(int id, int L, int R, int wz) {
        if(!id) return (ll)inf * inf;
        int mid = (L + R) >> 1;
        if(wz <= mid) return min(seg[id].get(wz), qry(ch[id][0], L, mid, wz));
        else return min(seg[id].get(wz), qry(ch[id][1], mid + 1, R, wz));   
    }
}lc;
int n, m, a[N], b[N], rt;
int main(){
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;++i){
        scanf("%d",&a[i]);
    }
    sort(a + 1, a + n + 1);
    reverse(a + 1, a + n + 1);
    for(int i=1;i<=n;++i){
        lc.add(rt, 0, inf, {-i, -(ll)a[i] * i});
    }
    for(int i=1;i<=m;++i) {
        scanf("%d",&b[i]);
        printf("%lld ",-lc.qry(rt, 0, inf, b[i]));
    }
    return 0;
}