P4655 [CEOI2017]Building Bridges 斜率优化

(Updated: )
(4 mins to read)

有n根柱子，第i根柱子的高度为$h_i$​，被拆除的代价为$w_i$​，在j和i两根柱子间架桥的费用为$(h_i-h_j)^2 + \sum \limits_{k=j+1}^{i-1} w_k$​，问连接1和n的最小代价

做法

$dp_i = \min \limits_{j<i} \{dp_j + (h_i-h_j)^2 + s_{i-1} - s_j\}$ 此处s为w的前缀和显然可以化为斜率优化的形式每次询问$h_i$​处y的最小值，然后插入一条斜率为$-2h_i$​，截距为$h_i^2-s_i+dp_i$​的直线。由于这里的$h_i$​不具备单调性，所以不能用队列来维护。可以用平衡树、李超树或者cdq分治解决

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#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int N = 1e5 + 5;
int n, h[N];
ll w[N], dp[N];
bool Cmp;
struct Line
{
  mutable ll k, b, p;
  bool operator < (const Line& oth) const { return Cmp ? p < oth.p : k < oth.k; }
};
struct ConvecHull : multiset<Line>
{
  static const ll inf = LLONG_MAX;
  ll div(ll a, ll b)
  {
    if(!b) return a>0 ? inf : -inf;
    return a / b - ((a ^ b) < 0 && a % b);
  }
  bool isect(iterator x, iterator y)
  {
    if (y == end()) { x->p = inf; return false; }
    if (x->k == y->k) x->p = x->b > y->b ? inf : -inf;
    else x->p = div((y->b - x->b), (x->k - y->k));
    return x->p >= y->p;
  }
  void add(ll k, ll b)
  {
    auto z = insert({k, b, 0}), y = z++, x = y;
    while (isect(y, z)) z = erase(z);
    if (x != begin() && isect(--x, y)) isect(x, y = erase(y));
    while ((y = x) != begin() && (--x)->p >= y->p) isect(x, erase(y));
  }
  ll qry(ll x)
  {
    Cmp = 1; auto l = *lower_bound({0, 0, x}); Cmp = 0;
    return l.k * x + l.b;
  }
}ch;
int main()
{
  scanf("%d", &n);
  for(int i=1; i<=n; i++) scanf("%d", h+i);
  for(int i=1; i<=n; i++) scanf("%lld", w+i), w[i] += w[i-1];
  ch.add(2*h[1], -1ll*h[1]*h[1]+w[1]);
  for(int i=2; i<=n; i++)
  {
    dp[i] = -ch.qry(h[i]) + w[i-1] + 1ll*h[i]*h[i];
    ch.add(2*h[i], -1ll*h[i]*h[i]+w[i]-dp[i]);
  }
  printf("%lld\n", dp[n]);
  return 0;
}