[NOI2007]货币兑换 「CDQ分治实现斜率优化」

首先每次买卖一定是在某天 $k$ 以当时的最大收入买入，再到第 $i$ 天卖出，那么易得方程：

$$f_i = \max \{\frac{A_iRate_kf_k}{A_kRate_k + B_k} + \frac{B_if_k}{A_kRate_k + B_k}\}$$

再令

$$\left\{\begin{aligned} x_k = \frac{Rate_kf_k}{A_kRate_k + B_k} \\ y_k = \frac{f_k}{A_kRate_k + B_k}\end{aligned}\right.$$

则有

$$\begin{aligned} f_i &= \max \{A_ix_k + B_iy_k\} \\ y_k &= - \frac{A_i}{B_i}x_k + \frac{f_i}{B_i} \end{aligned}$$

那么现在需要找到一个点 $(x_k, y_k)$ 使得直线的截距最大

由于斜率和横坐标皆不满足单调性，可以用平衡树等维护，这里使用CDQ分治实现

实现过程如下：

Ⅰ 将数据按照斜率$\frac{A_i}{B_i}$降序排序

Ⅱ 将区间按照操作顺序分为左右两部分处理

Ⅲ 先处理左半部分，维护左半边凸包（注意，此时左半边已按照 $x$ 排序）

Ⅳ 处理左半边对右半边的影响，由于已按照斜率降序排序，所以普通斜率优化即可

Ⅴ 将区间按照 $x$ 排序

代码

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <cmath>

 using namespace std;

 const int MAXN = 1e05 + ;

 const double INF = 1e60;
 const double eps = 1e-;

 int Dcmp (double p) {
     if (fabs (p) < eps)
         return ;
     return p <  ? -  : ;
 }

 struct CashSt {
     double a, b, rate;
     double k, x, y;
     int index;

     CashSt () {}

     bool operator < (const CashSt& p) const {
         return Dcmp (x - p.x) ==  ? Dcmp (y - p.y) <  : Dcmp (x - p.x) < ;
     }
 } ;
 CashSt Cash[MAXN];
 bool comp (const CashSt& a, const CashSt& b) {
     return Dcmp (a.k - b.k) > ;
 }

 int N;

 double slope (CashSt a, CashSt b) {
     if (Dcmp (b.x - a.x) == )
         return INF;
     return (b.y - a.y) / (b.x - a.x);
 }
 double f[MAXN]= {};
 CashSt Que[MAXN];
 int l = , r = ;
 CashSt temp[MAXN];
 void CDQ (int left, int right) {
     if (left == right) {
         f[left] = max (f[left], f[left - ]);
         Cash[left].y = f[left] / (Cash[left].a * Cash[left].rate + Cash[left].b);
         Cash[left].x = Cash[left].y * Cash[left].rate;
         return ;
     }
     int mid = (left + right) >> ;
     int p1 = left - , p2 = mid;
     for (int i = left; i <= right; i ++)
         Cash[i].index <= mid ? temp[++ p1] = Cash[i] : temp[++ p2] = Cash[i];
     for (int i = left; i <= right; i ++)
         Cash[i] = temp[i];
     CDQ (left, mid);
     l = , r = ;
     for (int i = left; i <= mid; i ++) {
         while (l < r && Dcmp (slope (Que[r - ], Que[r]) - slope (Que[r], Cash[i])) < )
             r --;
         Que[++ r] = Cash[i];
     }
     for (int i = mid + ; i <= right; i ++) {
         while (l < r && Dcmp (slope (Que[l], Que[l + ]) - Cash[i].k) > )
             l ++;
         f[Cash[i].index] = max (f[Cash[i].index], Cash[i].a * Que[l].x + Cash[i].b * Que[l].y);
     }
     CDQ (mid + , right);
     l = left, r = mid + ;
     int p = ;
     while (l <= mid && r <= right) {
         // if (Dcmp (Cash[l].x - Cash[r].x) < 0 || (Dcmp (Cash[l].x - Cash[r].x) == 0 && Dcmp (Cash[l].y - Cash[r].y) < 0))
         if (Cash[l] < Cash[r])
             temp[++ p] = Cash[l], l ++;
         else
             temp[++ p] = Cash[r], r ++;
     }
     while (l <= mid)
         temp[++ p] = Cash[l], l ++;
     while (r <= right)
         temp[++ p] = Cash[r], r ++;
     for (int i = ; i <= p; i ++)
         Cash[i + left - ] = temp[i];
 }

 double getnum () {
     double num = 0.0;
     char ch = getchar ();
     double T = 1.0;

     while (! isdigit (ch))
         ch = getchar ();
     while (isdigit (ch))
         num = num * 10.0 + (ch - '') * 1.0, ch = getchar ();
     if (ch == '.') {
         ch = getchar ();
         while (isdigit (ch))
             num = num + (T /= 10.0) * (ch - ''), ch = getchar ();
     }

     return num;
 }

 int main () {
     // freopen ("Input.txt", "r", stdin);

     scanf ("%d%lf", & N, & f[]);
     for (int i = ; i <= N; i ++) {
         double a = getnum (), b = getnum (), rate = getnum ();
         Cash[i].a = a, Cash[i].b = b, Cash[i].rate = rate;
         Cash[i].index = i;
         Cash[i].k = - a / b;
     }
     sort (Cash + , Cash + N + , comp);
     CDQ (, N);
     printf ("%.3f\n", f[N]);

     return ;
 }

 /*
 3 100
 1 1 1
 1 2 2
 2 2 3
 */