CF1083E The Fair Nut and Rectangles

CF1083E The Fair Nut and Rectangles

给定 \(n\) 个平面直角坐标系中左下角为坐标原点，右上角为 \((x_i,\ y_i)\) 的互不包含的矩形，每一个矩形拥有权值 \(a_i\) 。你的任务是选出若干个矩形，使得选出的矩形的面积并减去矩形的权值之和尽可能大。输出最大值。

\(1\leq n\leq10^6,\ 1\leq x_i,\ y_i\leq10^9,\ 0\leq a_i\leq x_i\cdot y_i\)

斜率优化

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因为所有矩阵互不包含，所以先按 \(x_i\) 升序， \(y_i\) 降序排个序

令 \(f_i\) 为，截止第 \(i\) 个矩阵，选取任意个矩阵所能得到的最大值

则： $$f_i=\displaystyle\max_{j<i}{f_j+S_i-S_i\cap S_j-a_i}$$

因为 \(x_i\) 升序， \(y_i\) 降序，所以 $$f_i=\displaystyle\max_{j<i}{f_j+x_iy_i+x_jy_i-a_i}$$

接着套路地化式子，得到 $$f_j=y_ix_j+f_i+a_i-x_iy_i$$

令 \(Y=f_j,\ K=y_i,\ X=x_j\) ，因为 \(y_i\) 单调递减，所以直接维护即可

时间复杂度 \(O(n)\)

代码

#include <bits/stdc++.h>
using namespace std;

#define nc getchar()
typedef long long ll;
typedef long double db;
const int maxn = 1e6 + 10;
ll f[maxn];
int n, q[maxn];

struct node {
  ll x, y, w;
  bool operator < (const node& o) const {
    return y > o.y;
  }
} a[maxn];

inline ll read() {
  ll x = 0;
  char c = nc;
  while (c < 48) c = nc;
  while (c > 47) x = (x << 3) + (x << 1) + (c ^ 48), c = nc;
  return x;
}

db slope(int x, int y) {
  return a[x].x == a[y].x ? 1e18 : db(f[x] - f[y]) / db(a[x].x - a[y].x);
}

int main() {
  n = read();
  for (int i = 1; i <= n; i++) {
    a[i].x = read();
    a[i].y = read();
    a[i].w = read();
  }
  ll ans = 0;
  int l = 1, r = 1;
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; i++) {
    while (l < r && slope(q[l], q[l + 1]) > a[i].y) l++;
    f[i] = f[q[l]] + (a[i].x - a[q[l]].x) * a[i].y - a[i].w;
    while (l < r && slope(q[r - 1], q[r]) < slope(q[r], i)) r--;
    q[++r] = i, ans = max(ans, f[i]);
  }
  printf("%I64d", ans);
  return 0;
}

orz \(\color{black}{S}\color{red}{ooke}\)