1652: [Usaco2006 Feb]Treats for the Cows
Time Limit: 5 Sec
Memory Limit: 64 MB
Submit: 275
Solved: 212
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Description
FJ has purchased N (1 <= N <= 2000) yummy treats for the cows who get money for giving vast amounts of milk. FJ sells one treat per day and wants to maximize the money he receives over a given period time. The treats are interesting for many reasons: * The treats are numbered 1..N and stored sequentially in single file in a long box that is open at both ends. On any day, FJ can retrieve one treat from either end of his stash of treats. * Like fine wines and delicious cheeses, the treats improve with age and command greater prices. * The treats are not uniform: some are better and have higher intrinsic value. Treat i has value v(i) (1 <= v(i) <= 1000). * Cows pay more for treats that have aged longer: a cow will pay v(i)*a for a treat of age a. Given the values v(i) of each of the treats lined up in order of the index i in their box, what is the greatest value FJ can receive for them if he orders their sale optimally? The first treat is sold on day 1 and has age a=1. Each subsequent day increases the age by 1.
约翰经常给产奶量高的奶牛发特殊津贴，于是很快奶牛们拥有了大笔不知该怎么花的钱．为此，约翰购置了N(1≤N≤2000)份美味的零食来卖给奶牛们．每天约翰售出一份零食．当然约翰希望这些零食全部售出后能得到最大的收益．这些零食有以下这些有趣的特性：
•零食按照1．．N编号，它们被排成一列放在一个很长的盒子里．盒子的两端都有开口，约翰每
天可以从盒子的任一端取出最外面的一个．
•与美酒与好吃的奶酪相似，这些零食储存得越久就越好吃．当然，这样约翰就可以把它们卖
出更高的价钱．
•每份零食的初始价值不一定相同．约翰进货时，第i份零食的初始价值为Vi(1≤Vi≤1000)．
•第i份零食如果在被买进后的第a天出售，则它的售价是vi×a．
Vi的是从盒子顶端往下的第i份零食的初始价值．约翰告诉了你所有零食的初始价值，并
希望你能帮他计算一下，在这些零食全被卖出后，他最多能得到多少钱．
Input
* Line 1: A single integer,
N * Lines 2..N+1: Line i+1 contains the value of treat v(i)
Output
* Line 1: The maximum revenue FJ can achieve by selling the treats
Sample Input
5
1
3
1
5
2
Five treats. On the first day FJ can sell either treat #1 (value 1) or
treat #5 (value 2).
Sample Output
43
OUTPUT DETAILS:
FJ sells the treats (values 1, 3, 1, 5, 2) in the following order
of indices: 1, 5, 2, 3, 4, making 1x1 + 2x2 + 3x3 + 4x1 + 5x5 = 43.
HINT
Source
题解：
dp.f[i][j]=max(f[i+1][j]+a[i]*t,f[i][j-1]+a[j]*t)
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#define ll long long
using namespace std;
int read()
{
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
int N,a[2005],f[2005][2005];
int main()
{
N=read();
for(int i=1;i<=N;i++)a[i]=read();
for(int k=1;k<=N;k++)
for(int i=1;i+k-1<=N;i++)
{
int t=N-k+1,j=i+k-1;
f[i][j]=max(f[i+1][j]+t*a[i],f[i][j-1]+t*a[j]);
}
cout<<f[1][N]<<endl;
}