题面
给你两个序列\(A, B\)， \(\forall u, v(u \not = v)\)之间边的权值为\(a_ua_v+b_ub_v\)。求最小生成树的边权和。
原题目
editorial
朴素的想法
考虑类似题目的做法，考虑每一次寻找最小的然后加入。发现这种思想和Boruvka比较相似。于是我们考虑Boruvka的方式来做。
对现有的连通块的基础上考虑：我们可以将这条新的边放在连通块编号在当前连通块前面的，也可以放在连通块编号在当前编号后面的（假设对当前的连通块标编号）。那么我们需要对这两部分分开计算。那么就可以转化问题了。
转化问题
问题变成了这样：
我们要维护一个数据结构，支持每次向集合\(S\)加入一个数对\((a, b)\), 并询问时给出\((x, y)\), 求\(\min \limits_{(a, b) \in S} (ax+by)\)
这个问题在maspy的题解里面写的是用凸包或CHT。
但是我不会geo的任何东西，于是转化成ds。
当\(b\not = 0\)时，相当不人类智慧的考虑到\(ax+by=y(\frac{x}{y}a+b)\)，看作时某个位置的一次函数最值乘上一个常数，发现可以直接李超线段树维护。
note
注意一下0的情况。至少我还需要除了线段树要计所有加入a的max，所有加入a的min，所有加入且b为0的a的max，所有加入且b为0的a的min，然后在做。详细的见代码（有点丑）
code
using db = double;
inline constexpr static db eps = 1e-10;constexpr int N = 5e4 + 5;int n, fa[N], dsu_con_cnt = 0;
ll ans = 0, val[N], a[N], b[N];
db pos[N];
vector<int> cons[N];
pii rcs[N];int find(int x) { return fa[x] == x ? x : fa[x] = find(fa[x]); }
bool merge(int u, int v) { auto du = find(u), dv = find(v); if (du != dv) return fa[du] = dv, --dsu_con_cnt, true; else return false; }struct Line {ll k, b;Line(ll _k = inf<ll>, ll _b = 0) : k(_k), b(_b) { }Line& operator=(const Line&x) { k = x.k, b = x.b; return *this; }
};struct segnode {Line l1, l2;
} seg[N << 2];# define lson index << 1
# define rson index << 1 | 1inline db calc(Line cur, db pos) {return static_cast<db>(cur.k) * pos + static_cast<db>(cur.b);
}void build(int index, int l, int r) { if (l > r) return ;seg[index].l1 = seg[index].l2 = Line(); if (l == r) return ;int mid = (l + r) >> 1;build(lson, l, mid);build(rson, mid + 1, r);
}void change1(int index, int l, int r, Line x) {if (l > r) return ;int mid = (l + r) >> 1;if (seg[index].l1.k == inf<ll> || calc(seg[index].l1, pos[mid]) < calc(x, pos[mid])) swap(seg[index].l1, x);if (x.k == inf<ll> || l == r) return ;if (calc(seg[index].l1, pos[l]) < calc(x, pos[l])) change1(lson, l, mid, x);if (calc(seg[index].l1, pos[r]) < calc(x, pos[r])) change1(rson, mid + 1, r, x);
}void query1(int index, int l, int r, int qpos, Line &cur) {if (l > r) return ;int mid = (l + r) >> 1;if (seg[index].l1.k == inf<ll>) return;if (cur.k == inf<ll> || calc(cur, pos[qpos]) < calc(seg[index].l1, pos[qpos])) cur = seg[index].l1;if (l == r) return ;if (qpos <= mid) query1(lson, l, mid, qpos, cur);else query1(rson, mid + 1, r, qpos, cur);
}void change2(int index, int l, int r, Line x) {if (l > r)return ;int mid = (l + r) >> 1;if (seg[index].l2.k == inf<ll> || calc(seg[index].l2, pos[mid]) > calc(x, pos[mid])) swap(seg[index].l2, x);if (x.k == inf<ll> || l == r) return ;if (calc(seg[index].l2, pos[l]) > calc(x, pos[l])) change2(lson, l, mid, x);if (calc(seg[index].l2, pos[r]) > calc(x, pos[r])) change2(rson, mid + 1, r, x);
}void query2(int index, int l, int r, int qpos, Line &cur) {if (l > r) return ;int mid = (l + r) >> 1;if (seg[index].l2.k == inf<ll>) return;if (cur.k == inf<ll> || calc(cur, pos[qpos]) > calc(seg[index].l2, pos[qpos])) cur = seg[index].l2;if (l == r) return ;if (qpos <= mid) query2(lson, l, mid, qpos, cur);else query2(rson, mid + 1, r, qpos, cur);
}struct segtree_cht {int sz, tot;bool seg1_added;map<pair<ll, ll>, int> idx;pair<ll, int> amx, amn, amx_zero, amn_zero;segtree_cht() { amx = mkp(-inf<ll>, -1), amn = mkp(inf<ll>, -1), amx_zero = mkp(-inf<ll>, -1), amn_zero = mkp(inf<ll>, -1); seg1_added = false; }void rsz(int _sz) { sz = _sz; }void work(ll *X, ll *Y) { rep(i, 1, n) idx[mkp(X[i], Y[i])] = i; rep(i, 1, n) if (Y[i]) ::pos[++tot] = db(X[i]) / db(Y[i]);if (tot) {sort(pos + 1, pos + tot + 1);tot = unique(pos + 1, pos + tot + 1) - pos - 1;}}void rst() { seg1_added = false; build(1, 1, tot); }void add(ll a, ll b) {auto _id = idx[mkp(a, b)];chkmax(amx, mkp(a, _id)), chkmin(amn, mkp(a, _id));if (!b) {auto _id = idx[mkp(a, b)];chkmax(amx_zero, mkp(a, _id)), chkmin(amn_zero, mkp(a, _id));}change1(1, 1, tot, Line(a, b));seg1_added = true;change2(1, 1, tot, Line(a, b));}ll query_max(ll x, ll y) {if (!seg1_added) return -inf<ll>;if (!y) {if (x >= 0) return x * amx.first;else return x * amn.first;}int p = lower_bound(pos + 1, pos + tot + 1, db(x) / y) - pos;Line tans;tans.k = inf<ll>;if (y >= 0) query1(1, 1, tot, p, tans);else query2(1, 1, tot, p, tans);return tans.k * x + tans.b * y;}pair<ll, int> query_max_with_id(ll x, ll y) {if (!seg1_added) return mkp(-inf<ll>, -1);if (!y) {if (x >= 0) return mkp(x * amx.first, amx.second);else return mkp(x * amn.first, amn.second);}int p = lower_bound(pos + 1, pos + tot + 1, db(x) / y) - pos;Line tans;tans.k = inf<ll>;if (y > 0) query1(1, 1, tot, p, tans);else query2(1, 1, tot, p, tans);pair<ll, int> curans = mkp(tans.k * x + tans.b * y, idx[mkp(tans.k, tans.b)]);if (x >= 0) {if (amx_zero.first != -inf<ll>) chkmax(curans, mkp(x * amx_zero.first, amx_zero.second));} else {if (amn_zero.first != inf<ll>) chkmax(curans, mkp(x * amn_zero.first, amn_zero.second));}return curans;}ll query_min(ll x, ll y) {return -query_max(-x, -y);}pair<ll, int> query_min_with_id(ll x, ll y) {auto ret = query_max_with_id(-x, -y);return mkp(-ret.first, ret.second);}
};segtree_cht T;signed main() { read(n); T.rsz(n);iota(fa + 1, fa + n + 1, 1);dsu_con_cnt = n;rep(i, 1, n) read(a[i]);rep(i, 1, n) read(b[i]);T.work(a, b);while(dsu_con_cnt > 1) {bool flag = false;rep(i, 1, n) cons[i].clear();rep(i, 1, n) cons[find(i)].eb(i);fill(val + 1, val + n + 1, inf<ll>);fill(rcs + 1, rcs + n + 1, mkp(0, 0));T.rst();rep(u, 1, n) {for(auto v : cons[u]) {auto [tv, w] = T.query_min_with_id(a[v], b[v]);// dbg(u, v, w, tv);if (val[u] > tv) val[u] = tv, rcs[u] = mkp(v, w);}for (auto v : cons[u]) T.add(a[v], b[v]);}T.rst();per(u, n, 1) {for(auto v : cons[u]) {auto [tv, w] = T.query_min_with_id(a[v], b[v]);// dbg(u, v, w, tv);if (val[u] > tv) val[u] = tv, rcs[u] = mkp(v, w);} for (auto v : cons[u]) T.add(a[v], b[v]);}rep(i, 1, n) {auto [u, v] = rcs[i];if (!u) continue;if (merge(u, v)) {flag = true;ans += val[i];}}if (!flag) break;}writeln(ans);#if defined(LOCAL) && !defined(CPH)std::cerr << "Spend Time : " << clock() * 1. / CLOCKS_PER_SEC * 1e3 << " ms \n";
#endifreturn 0;
}
后来发现我的想法和maspy的基本相同，而李超树写法好像比maspy的CHT板子跑的快！
