## 1492: [NOI2007] Currency conversion Cash

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 3396  Solved: 1434
[Submit][Status][Discuss]

## Description

Small Y Recently I worked in a stock exchange . The exchange issues and trades only two types of securities ：A Souvenir Ticket （ hereinafter referred to as A Coupon ） and B Souvenir Ticket （ following
abbreviation B Coupon ）. Every customer with a coupon has his own account . The number of coupons can be a real number . Every day with the ups and downs of the market ,
Both have their own value at the time , That is, the amount of RMB that can be exchanged for each unit of coupon on the same day . We record the number of K In the day A Coupon and B Coupon Of
The values are AK and BK（ element / Unit coupon ）. For the convenience of customers , The stock exchange provides a very convenient way of trading ： The proportional trading method
. The proportional trading method is divided into two aspects ：（a） Sell coupons ： The customer offers a [0,100] Real number in OP As a percentage of sales , Its significance is ： take
OP% Of A Coupons and OP% Of B Coupon At that time the value of RMB ;（b） Buy the coupons ： Customers pay IP RMB , The exchange will exchange
The total value to users is IP The gold coupon for , also , To satisfy what is offered to customers A Coupons and B The proportion of coupons is in K It happened that RateK; for example , Suppose that
Come down 3 Within days Ak、Bk、RateK The changes are as follows ： Suppose on the first day , Users have 100 element RMB, but without any vouchers . The user can do the following ： be aware , It can be operated many times in the same day . Small Y He is a very economical employee , Through a long period of operation and market estimation , He has
Knowing the future N Within days A Coupons and B The value of the voucher and Rate. He also wants to be able to work it out , If you start with S Yuan , that N At most in a few days
How much is enough .

## Input

Enter two positive integers on the first line N、S, They are small Y The number of days you can predict and the amount of money you have initially . Next N That's ok , The first K Row three real numbers AK、B
K、RateK, The meaning is described in the title . about 100% Test data for , Satisfy ：0<AK≤10;0<BK≤10;0<RateK≤100;MaxProfit≤1
0^9.
【 Tips 】
1. The input file can be large , Please use a quick read in method .
2. There must be an optimal trading scheme to satisfy ：
Each buying operation USES up all the yuan ;
Every time I sell, I sell all of my gold notes .

## Output

There's only one real number MaxProfit, It means the first one N The maximum amount of money you can get at the end of a day's operation . The answer is reserved 3 Decimal place .

3 100
1 1 1
1 2 2
2 2 3

225.000

## HINT ## Solution

Slope optimization DP, Typical is not monotonous

$f[i]=max（f[i],f[j]/(a[j]*rate[j]+b[j])*rate[j]*a[i]+f[j]/(a[j]*rate[j]+b[j])*b[i]）$

as for CDQ The theory of partition ： Walk   Over and over

inspire ：

1. Slope optimization DP There are thousands of changes ..

2. Mastering the divide and conquer algorithm can produce many magical effects （xyx I seem to like divide and conquer very much ）

## Code

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
#define maxn 200010
struct DayNode
{
double A,B,Ra,k; int id;
bool operator < (const DayNode & T) const
{return k<T.k;}
}a[maxn],tmpa[maxn];
struct PointNode{double x,y;}p[maxn],tmpp[maxn];
int N;
double dp[maxn];
int tmp1[maxn],tmp2[maxn],stack[maxn];
#define inf 1e9
#define eps 1e-9
double slope(int l1,int l2)
{
if (!l1) return -inf;
if (!l2) return inf;
if (fabs(p[l1].x-p[l2].x)<=eps) return -inf;
return (p[l1].y-p[l2].y)/(p[l1].x-p[l2].x);
}
void CDQ(int l,int r)
{
if (l==r)
{
dp[l]=max(dp[l],dp[l-]);
p[l].y=dp[l]/(a[l].Ra*a[l].A+a[l].B);
p[l].x=a[l].Ra*p[l].y;
return;
}
int mid=(l+r)>>,L,R;
L=l; R=mid+;
for (int i=l; i<=r; i++)
if (a[i].id<=mid) tmpa[L++]=a[i]; else tmpa[R++]=a[i];
for (int i=l; i<=r; i++) a[i]=tmpa[i];
CDQ(l,mid);
int top=;
for (int i=l; i<=mid; i++)
{
while (top> && slope(stack[top],stack[top-])<slope(i,stack[top-])+eps) top--;
stack[++top]=i;
}
int Top=;
for (int i=r; i>=mid+; i--)
{
while (Top<top && slope(stack[Top],stack[Top+])+eps>a[i].k) Top++;
dp[a[i].id]=max(dp[a[i].id],p[stack[Top]].x*a[i].A+p[stack[Top]].y*a[i].B);
}
CDQ(mid+,r);
L=l; R=mid+;
for (int i=l; i<=r; i++)
if (((p[L].x<p[R].x+eps || (fabs(p[L].x-p[R].x)<=eps && p[L].y<p[R].y+eps)) || R>r) && L<=mid)
tmpp[i]=p[L++];
else tmpp[i]=p[R++];
for (int i=l; i<=r; i++) p[i]=tmpp[i];
}
int main()
{
scanf("%d%lf",&N,&dp[]);
for (int i=; i<=N; i++)
scanf("%lf%lf%lf",&a[i].A,&a[i].B,&a[i].Ra),a[i].k=-a[i].A/a[i].B,a[i].id=i;
sort(a+,a+N+);
CDQ(,N);
//for (int i=1; i<=N; i++) printf("%d  %d  %.3lf  %.3lf  %.3lf  %.3lf  \n",i,a[i].id,a[i].A,a[i].B,a[i].Ra,a[i].k);
printf("%.3lf\n",dp[N]);
return ;
}

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