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。
(1leq nleq10^6, 1leq x_i, y_ileq10^9, 0leq a_ileq x_icdot y_i)
斜率优化
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: $$f_i=displaystylemax_{j<i}{f_j+S_i-S_icap S_j-a_i}$$
因为 (x_i) 升序, (y_i) 降序,所以 $$f_i=displaystylemax_{j<i}{f_j+x_iy_i+x_jy_i-a_i}$$
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})